case study questions on straight lines class 11

CBSE Case Study Questions for Class 11 Maths Straight Lines Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 11 Maths Straight Lines  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 11 Maths Straight Lines PDF

Checkout our case study questions for other chapters.

• Chapter 8 Binomial Theorem Case Study Questions
• Chapter 9 Sequences and Series Case Study Questions
• Chapter 11 Conic Sections Case Study Questions
• Chapter 12 Introduction to 3D Geometry Case Study Questions

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Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

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Class 11 Mathematics Case Study Questions

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If you’re seeking a comprehensive and dependable study resource with Class 11 mathematics case study questions for CBSE, myCBSEguide is the place to be. It has a wide range of study notes, case study questions, previous year question papers, and practice questions to help you ace your examinations. Furthermore, it is routinely updated to bring you up to speed with the newest CBSE syllabus. So, why delay? Begin your path to success with myCBSEguide now!

The rationale behind teaching Mathematics

The general rationale to teach Mathematics at the senior secondary level is to assist students:

• In knowledge acquisition and cognitive understanding of basic ideas, words, principles, symbols, and mastery of underlying processes and abilities, notably through motivation and visualization.
• To experience the flow of arguments while demonstrating a point or addressing an issue.
• To use the information and skills gained to address issues using several methods wherever possible.
• To cultivate a good mentality in order to think, evaluate, and explain coherently.
• To spark interest in the subject by taking part in relevant tournaments.
• To familiarise pupils with many areas of mathematics utilized in daily life.
• To pique students’ interest in studying mathematics as a discipline.

Case studies in Class 11 Mathematics

A case study in mathematics is a comprehensive examination of a specific mathematical topic or scenario. Case studies are frequently used to investigate the link between theory and practise, as well as the connections between different fields of mathematics. A case study will frequently focus on a specific topic or circumstance and will investigate it using a range of methodologies. These approaches may incorporate algebraic, geometric, and/or statistical analysis.

Sample Class 11 Mathematics case study questions

When it comes to preparing for Class 11 Mathematics, one of the best things Class 11 Mathematics students can do is to look at some Class 11 Mathematics sample case study questions. Class 11 Mathematics sample case study questions will give you a good idea of the types of Class 11 Mathematics sample case study questions that will be asked in the exam and help you to prepare more effectively.

Looking at sample questions is also a good way to identify any areas of weakness in your knowledge. If you find that you struggle with a particular topic, you can then focus your revision on that area.

myCBSEguide offers ample Class 11 Mathematics case study questions, so there is no excuse. With a little bit of preparation, Class 11 Mathematics students can boost their chances of getting the grade they deserve.

Some samples of Class 11 Mathematics case study questions are as follows:

Class 11 Mathematics case study question 1

• 9 km and 13 km
• 9.8 km and 13.8 km
• 9.5 km and 13.5 km
• 10 km and 14 km
• x  ≤   −1913
• x <  −1613
• −1613  < x <  −1913
• There are no solution.
• y  ≤   12 x+2
• y >  12 x+2
• y  ≥   12 x+2
• y <  12 x+2

Answer Key:

• (b) 9.8 km and 13.8 km
• (a) −1913   ≤  x 
• (b)  y >  12 x+2
• (d) (-5, 5)

Class 11 Mathematics case study question 2

• 2 C 1 × 13 C 10
• 2 C 1 × 10 C 13
• 1 C 2 × 13 C 10
• 2 C 10 × 13 C 10
• 6 C 2​ × 3 C 4   × 11 C 5 ​
• 6 C 2​ × 3 C 4   × 11 C 5
• 6 C 2​ × 3 C 5 × 11 C 4 ​
• 6 C 2 ​  ×   3 C 1 ​  × 11 C 5 ​
• (b) (13) 4  ways
• (c) 2860 ways.

Class 11 Mathematics case study question 3

Read the Case study given below and attempt any 4 sub parts: Father of Ashok is a builder, He planned a 12 story building in Gurgaon sector 5. For this, he bought a plot of 500 square yards at the rate of Rs 1000 /yard². The builder planned ground floor of 5 m height, first floor of 4.75 m and so on each floor is 0.25 m less than its previous floor.

Class 11 Mathematics case study question 4

Read the Case study given below and attempt any 4 sub parts: villages of Shanu and Arun’s are 50km apart and are situated on Delhi Agra highway as shown in the following picture. Another highway YY’ crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at pints A and B such that OA=10 km and OB =12 km. Also, the villages of Barun and Jeetu are on the smaller high way YY’. Barun’s village B is 12km from O and that of Jeetu is 15 km from O.

Now answer the following questions:

• 5x + 6y = 60
• 6x + 5y = 60
• (a) (10, 0)
• (b) 6x + 5y = 60
• (b) 60/√ 61 km
• (d) 2√61 km

A peek at the Class 11 Mathematics curriculum

The Mathematics Syllabus has evolved over time in response to the subject’s expansion and developing societal requirements. The Senior Secondary stage serves as a springboard for students to pursue higher academic education in Mathematics or professional subjects such as Engineering, Physical and Biological Science, Commerce, or Computer Applications. The current updated curriculum has been prepared in compliance with the National Curriculum Framework 2005 and the instructions provided by the Focus Group on Teaching Mathematics 2005 in order to satisfy the rising demands of all student groups. Greater focus has been placed on the application of various principles by motivating the themes from real-life events and other subject areas.

Class 11 Mathematics (Code No. 041)

Design of Class 11 Mathematics exam paper

CBSE Class 11 mathematics question paper is designed to assess students’ understanding of the subject’s essential concepts. Class 11 mathematics question paper will assess their problem-solving and analytical abilities. Before beginning their test preparations, students in Class 11 maths should properly review the question paper format. This will assist Class 11 mathematics students in better understanding the paper and achieving optimum scores. Refer to the Class 11 Mathematics question paper design provided.

 Class 11 Mathematics Question Paper Design

• No chapter-wise weightage. Care to be taken to cover all the chapters.
• Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.  

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections.

  Prescribed Books:

• Mathematics Textbook for Class XI, NCERT Publications
• Mathematics Exemplar Problem for Class XI, Published by NCERT
• Mathematics Lab Manual class XI, published by NCERT

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CBSE Class 11 Maths – Chapter 10 Straight Lines – Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Straight Lines : Notes and Study Materials -pdf

• Concepts of  Straight Lines
• Straight Lines Master File
• Straight Lines Revision Notes
• R D Sharma Solution of Straight Lines
• R D Sharma Solution of Cartesian System
• NCERT Solution  Straight Lines
• NCERT  Exemplar Solution Straight Lines
• Straight Lines : Solved Example 1

CBSE Class 11 Maths Notes Chapter 10 Straight Lines

Distance Formula The distance between two points A(x 1 , y 1 ) and B (x 2 , y 2 ) is given by

The distance of a point A(x, y) from the origin 0 (0, 0) is given by OA =  x 2 + y 2 −−−−−−√

Section Formula The coordinates of the point which divides the joint of (x 1 , y 1 ) and (x 2 , y 2 ) in the ratio m : n internally, is

Mid-point of the joint of (x 1 , y 1 ) and (x 2 , y 2 ) is

X-axis divides the line segment joining (x 1 , y 1 ) and (x 2 , y 2 ) in the ratio -y 1  : y 2 .

Y-axis divides the line segment joining (x 1 , y 1 ) and (x 2 , y 2 ) in the ratio -x 1  : x 2 .

The coordinates of the centroid of the triangle whose vertices are (x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 ) is

Area of Triangle The area of the triangle, the coordinates of whose vertices are (x 1 , y 1 ), (x 2 , y 2 )and (x 3 , y 3 ) is the absolute value of

If the points (x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 ) are collinear, then x 1  (y 2  – y 3 ) + x 2  (y 3  – y 1 ) + x 3  (y 1  – y 2 ) = 0.

Shifting of Origin Let the origin is shifted to a point O'(h, k). If P(x, y) are coordinates of a point referred to old axes and P'(X, Y) are the coordinates of the same points referred to new axes, then x = X + h, y = Y + k.

Straight Line Any curve is said to be a straight line if two points are taken on the curve such that every point on the line segment joining any two points on it lies on the curve. General equation of a line is ax + by + c = 0.

Slope or Gradient of Line The inclination of angle θ to a line with a positive direction of X-axis in the anti-clockwise direction, the tangent of angle θ is said to be slope or gradient of the line and is denoted by m. i.e. m = tan θ The slope of a line passing through points P(x 1 , y 1 ) and Q(x 2 , y 2 ) is given by

Note:  Slope of a line parallel to X-axis is zero and slope of a line parallel to Y-axis is not defined.

Angle between Two Lines The angle θ between two lines having slope m 1  and m 2  is

• If two lines are parallel, their slopes are equal i.e. m 1  = m 2 .
• If two lines are perpendicular to each other, then their product of slopes is -1 i.e. m 1 m 2  = -1.

Various Forms of the Equation of a Line If a line is at a distance k and parallel to X-axis, then the equation of the line is y = ± k.

If a line is parallel to Y-axis at a distance c from Y-axis, then its equation is x = ± c.

Slope-intercept form: The equation of line with slope m and making an intercept c on the y-axis, is y = mx + c.

One point-slope form: The equation of a line which passes through the point (x 1 , y 1 ) and has the slope of m is given by y – y 1  = m (x – x 1 ).

Two points form: The equation of a line passing through the points (x 1 , y 1 ) and (x 2 , y 2 ) is given by

The Intercept form: The equation of a line which cuts off intercepts a and b respectively on the x and y-axes is given by  x a + y b = 1

The normal form: The equation of a straight line upon which the length of the perpendicular from the origin is p and angle made by this perpendicular to the x-axis is α, is given by x cos α + y sin α = p.

General Equation of a Line Any equation of the form Ax + By + C = 0, where A and B are simultaneously not zero is called the general equation of a line.

Different Forms of Ax + By + C = 0 Slope intercept form: If B ≠ 0, then Ax + By + C = 0 can be written as

If B = 0, then x = – C / A which is a vertical line, whose slope is not defined and x-intercept is – C/A.

Intercept form If C ≠ 0, then Ax + By + C = 0 can be written as

where a = – C / A and b = – C/B If C = 0, then Ax + By + C = 0 can be written as Ax + By = 0 which is a line passing through origin and therefore has zero intercept on the axes.

Normal form: The normal form of equation Ax + By + C = 0 is x cos α + y sin α = p where

Note: Proper choice of signs to be made so that p should be always positive.

Position of Points is Relative to a Given Line Let the equation of the given line be ax + by + c = 0 and let the coordinates of the two given points be P(x 1 , y 1 ) and Q(x 2 , y 2 ). The two points are on the same side of the straight line ax + by + c = 0, If ax 1  + by 1  + c and ax 2  + by 2  + c have the same sign.

The two points are on the opposite sides of the straight line ax + by + c = 0, If ax 1  + by 1  + c and ax 2  + by 2  + c have opposite sign.

Condition of concurrency for three given lines a 1 x + b 1 y + c 1  = 0, a 2 x + b 2 y + c 2  = 0 and a 3 x + b 3 y+ c 3  = 0 is a 3 (b 1 c 2  – b 2 c 1 ) + b 3 (a 2 c 1  – a 1 c 2 ) + c 3 (a 1 b 2  – a 2 b 1 ) = 0

Point of intersection of two lines Let equation of lines be ax 1  + by 1  + c 1  = 0 and a 2 x + b 2 y + c 2 = 0, then their point of intersection is

Distance of a Point from a Line The perpendicular distanced of a point P(x 1 , y 1 )from the line Ax + By + C = 0 is given by

Distance Between Two Parallel Lines The distance d between two parallel lines y = mx + c 1  and y = mx + c 2  is given by

Straight Lines Class 11 MCQs Questions with Answers

Question 1. The locus of a point, whose abscissa and ordinate are always equal is (a) x + y + 1 = 0 (b) x – y = 0 (c) x + y = 1 (d) none of these.

Answer: (b) x – y = 0 Hint: Let the coordinate of the variable point P is (x, y) Now, the abscissa of this point = x and its ordinate = y Given, abscissa = ordinate ⇒ x = y ⇒ x – y = 0 So, the locus of the point is x – y = 0

Question 2. The equation of straight line passing through the point (1, 2) and parallel to the line y = 3x + 1 is (a) y + 2 = x + 1 (b) y + 2 = 3 × (x + 1) (c) y – 2 = 3 × (x – 1) (d) y – 2 = x – 1

Answer: (c) y – 2 = 3 × (x – 1) Hint: Given straight line is: y = 3x + 1 Slope = 3 Now, required line is parallel to this line. So, slope = 3 Hence, the line is y – 2 = 3 × (x – 1)

Question 3. What can be said regarding if a line if its slope is negative (a) θ is an acute angle (b) θ is an obtuse angle (c) Either the line is x-axis or it is parallel to the x-axis. (d) None of these

Answer: (b) θ is an obtuse angle Hint: Let θ be the angle of inclination of the given line with the positive direction of x-axis in the anticlockwise sense. Then its slope is given by m = tan θ Given, slope is positive ⇒ tan θ < 0 ⇒ θ lies between 0 and 180 degree ⇒ θ is an obtuse angle

Question 4: The equation of the line which cuts off equal and positive intercepts from the axes and passes through the point (α, β) is (a) x + y = α + β (b) x + y = α (c) x + y = β (d) None of these

Answer: (a) x + y = α + β Hint: Let the equation of the line be x/a + y/b = 1 which cuts off intercepts a and b with the coordinate axes. It is given that a = b, therefore the equation of the line is x/a + y/a = 1 ⇒ x + y = a …..1 But it is passes through (α, β) So, α + β = a Put this value in equation 1, we get x + y = α + β

Question 5. Two lines a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 are coincedent if (a) a 1 /a 2 = b 1 /b 2 ≠ c 1 /c 2 (b) a 1 /a 2 ≠ b 1 /b 2 = c 1 /c 2 (c) a 1 /a 2 ≠ b 1 /b 2 ≠ c 1 /c 2 (d) a 1 /a 2 = b 1 /b 2 = c 1 /c 2

Answer: (d) a 1 /a 2 = b 1 /b 2 = c 1 /c 2 Hint: Two lines a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 are coincedent if a 1 /a 2 = b 1 /b 2 = c 1 /c 2

Question 6: The equation of the line passing through the point (2, 3) with slope 2 is (a) 2x + y – 1 = 0 (b) 2x – y + 1 = 0 (c) 2x – y – 1 = 0 (d) 2x + y + 1 = 0

Answer: (c) 2x – y – 1 = 0 Hint: Given, the point (2, 3) and slope of the line is 2 By, slope-intercept formula, y – 3 = 2(x – 2) ⇒ y – 3 = 2x – 4 ⇒ 2x – 4 – y + 3 = 0 ⇒ 2x – y – 1 = 0

Question 7. The slope of the line ax + by + c = 0 is (a) a/b (b) -a/b (c) -c/b (d) c/b

Answer: (b) -a/b Hint: Give, equation of line is ax + by + c = 0 ⇒ by = -ax – c ⇒ y = (-a/b)x – c/b It is in the form of y = mx + c Now, slope m = -a/b

Question 8. Equation of the line passing through (0, 0) and slope m is (a) y = mx + c (b) x = my + c (c) y = mx (d) x = my

Answer: (c) y = mx Hint: Equation of the line passing through (x 1 , y 1 ) and slope m is (y – y 1 ) = m(x – x 1 ) Now, required line is (y – 0 ) = m(x – 0) ⇒ y = mx

Question 9. The angle between the lines x – 2y = y and y – 2x = 5 is (a) tan -1 (1/4) (b) tan -1 (3/5) (c) tan -1 (5/4) (d) tan -1 (2/3)

Answer: (c) tan -1 (5/4) Hint: Given, lines are: x – 2y = 5 ………. 1 and y – 2x = 5 ………. 2 From equation 1, x – 5 = 2y ⇒ y = x/2 – 5/2 Here, m 1 = 1/2 From equation 2, y = 2x + 5 Here. m 2 = 2 Now, tan θ = |(m 1 + m 2 )/{1 + m 1 × m 2 }| = |(1/2 + 2)/{1 + (1/2) × 2}| = |(5/2)/(1 + 1)| = |(5/2)/2| = 5/4 ⇒ θ = tan -1 (5/4)

Question 10. Two lines a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 are parallel if (a) a 1 /a 2 = b 1 /b 2 ≠ c 1 /c 2 (b) a 1 /a 2 ≠ b 1 /b 2 = c 1 /c 2 (c) a 1 /a 2 ≠ b 1 /b 2 ≠ c 1 /c 2 (d) a 1 /a 2 = b 1 /b 2 = c 1 /c 2

Answer: (a) a 1 /a 2 = b 1 /b 2 ≠ c 1 /c 2 Hint: Two lines a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 are parallel if a 1 /a 2 = b 1 /b 2 ≠ c 1 /c 2

Question 11. The locus of a point, whose abscissa and ordinate are always equal is (a) x + y + 1 = 0 (b) x – y = 0 (c) x + y = 1 (d) none of these.

Question 12. In a ΔABC, if A is the point (1, 2) and equations of the median through B and C are respectively x + y = 5 and x = 4, then B is (a) (1, 4) (b) (7, – 2) (c) none of these (d) (4, 1)

Answer: (b) (7, – 2) Hint: The equation of median through B is x + y = 5 The point B lies on it. Let the coordinates of B are (x 1 , 5 – x 1 ) Now CF is a median through C, So co-ordiantes of F i.e. mid-point of AB are ((x 1 +1)/2, (5 – x 1 + 2)/2) Now since this lies on x = 4 ⇒ (x 1 + 1)/2 = 4 ⇒ x 1 + 1 = 8 ⇒ x 1 = 7 Hence, the co-oridnates of B are (7, -2)

Question 13. The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150 degrees with the positive direction of the y-axis. Then the equation of line is (a) x + y = 14 (b) √3y + x = 14 (c) √3x + y = 14 (d) None of these

Answer: (c) √3x + y = 14 Hint: Given, The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150 degrees with the positive direction of the y-axis. Now, equation of line is x × cos 30 + y × sin 30 = 7 ⇒ √3x/2 + y/2 = 7 ⇒ √3x + y = 7×2 ⇒ √3x + y = 14

Question 14. If two vertices of a triangle are (3, -2) and (-2, 3) and its orthocenter is (-6, 1) then its third vertex is (a) (5, 3) (b) (-5, 3) (c) (5, -3) (d) (-5, -3)

Answer: (d) (-5, -3) Hint: Let the third vertex of the triangle is C(x, y) Given, two vertices of a triangle are A(3,-2) and B(-2,3) Now given orthocentre of the circle = H(-6, 1) So, AH ⊥ BC and BH ⊥ AC Since the product of the slope of perpendicular lines equal to -1 Now, AH ⊥ BC ⇒ {(-2 – 1)/(3 + 6)} × {(y + 2)/(x – 3)} = -1 ⇒ (-3/9) × {(y + 2)/(x – 3)} = -1 ⇒ (-1/3)×{(y – 3)/(x + 2)} = -1 ⇒ (y – 3)/{3×(x + 2)} = 1 ⇒ (y – 3) = 3×(x + 2) ⇒ y – 3 = 3x + 6 ⇒ 3x + 6 – y = -3 ⇒ 3x – y = -3 – 6 ⇒ 3x – 2y = -9 ………… 1 Again, BH ⊥ AC ⇒ {(3 – 1)/(-2 + 6)} × {(y – 3)/(x + 2)} = -1 ⇒ (2/4) × {(y – 3)/(x + 2)} = -1 ⇒ (1/2)×{(y – 3)/(x + 2)} = -1 ⇒ (y – 3)/{2×(x + 2)} = 1 ⇒ (y – 3) = 2×(x + 2) ⇒ y – 3 = 2x + 4 ⇒ 2x + 4 – y = -3 ⇒ 2x – y = -3 – 4 ⇒ 2x – y = -7 ………… 2 Multiply equation 2 by 2, we get 4x – 2y = -14 ……… 3 Subtract equation 1 and we get -x = 5 ⇒ x = -5 From equation 2, we get 2×(-5) – y = -7 ⇒ -10 – y = -7 ⇒ y = -10 + 7 ⇒ y = -3 So, the third vertex of the triangle is (-5, -3)

Question 15. The sum of squares of the distances of a moving point from two fixed points (a, 0) and (-a, 0) is equal to 2c² then the equation of its locus is (a) x² – y² = c² – a² (b) x² – y² = c² + a² (c) x² + y² = c² – a² (d) x² + y² = c² + a²

Answer: (c) x² + y² = c² – a² Hint: Let P(h, k) be any position of the moving point and let A(a, 0) and B(-a, 0) be the given points. Then PA² + PB² = 2c² ⇒ (h – a)² + (k – 0)² + (h + a)² + (k – 0)² = 2c² ⇒ h² – 2ah + a² + k² + h² + 2ah + a² + k² = 2c² ⇒ 2h² + 2k² + 2a² = 2c² ⇒ h² + k² + a² = c² ⇒ h² + k² = c² – a² Hence, the locus of (h, k) is x² + y² = c² – a²

Question 16. The equation of the line through the points (1, 5) and (2, 3) is (a) 2x – y – 7 = 0 (b) 2x + y + 7 = 0 (c) 2x + y – 7 = 0 (d) x + 2y – 7 = 0

Answer: (c) 2x + y – 7 = 0 Hint: Given, points are: (1, 5) and (2, 3) Now, equation of line is y – y 1 = {(y 2 – y 1 )/(x 2 – x 1 )} × (x – x 1 ) ⇒ y – 5 = {(3 – 5)/(2 – 1)} × (x – 1) ⇒ y – 5 = (-2) × (x – 1) ⇒ y – 5 = -2x + 2 ⇒ 2x + y – 5 – 2 = 0 ⇒ 2x + y – 7 = 0

Question 17. What can be said regarding if a line if its slope is zero (a) θ is an acute angle (b) θ is an obtuse angle (c) Either the line is x-axis or it is parallel to the x-axis. (d) None of these

Answer: (c) Either the line is x-axis or it is parallel to the x-axis. Hint: Let θ be the angle of inclination of the given line with the positive direction of x- axis in the anticlockwise sense. Then its slope is given by m = tan θ Given, slope is zero ⇒ tan θ = 0 ⇒ θ = 0° ⇒ Either the line is x-axis or it is parallel to the x-axis.

Question 18. Two lines are perpendicular if the product of their slopes is (a) 0 (b) 1 (c) -1 (d) None of these

Answer: (c) -1 Hint: Let m 1 is the slope of first line and m 2 is the slope of second line. Now, two lines are perpendicular if m 1 × m 2 = -1 i.e. the product of their slopes is equals to -1

Question 19. y-intercept of the line 4x – 3y + 15 = 0 is (a) -15/4 (b) 15/4 (c) -5 (d) 5

Answer: (d) 5 Hint: Given, equation of line is 4x – 3y + 15 = 0 ⇒ 4x – 3y = -15 ⇒ 4x/(-15) + (-3)y/(-15) = 1 ⇒ x/(-15/4) + 3y/15 = 1 ⇒ x/(-15/4) + y/(15/3) = 1 ⇒ x/(-15/4) + y/5 = 1 Now, compare with x/a + y/b = 1, we get y-intercept b = 5

Question 20. The equation of the locus of a point equidistant from the point A(1, 3) and B(-2, 1) is (a) 6x – 4y = 5 (b) 6x + 4y = 5 (c) 6x + 4y = 7 (d) 6x – 4y = 7

Answer: (b) 6x + 4y = 5 Hint: Let P(h, k) be any point on the locus. Then Given, PA = PB ⇒ PA² = PB² ⇒ (h – 1)² + (k – 3)² = (h + 2)² + (k – 1)² ⇒ h² – 2h + 1 + k² – 6k + 9 = h² + 4h + 4 + k² – 2k + 1 ⇒ -2h – 6k + 10 = 4h – 2k + 5 ⇒ 6h + 4k = 5 Hence, the locus of (h, k) is 6x + 4y = 5

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• Important Questions for Class 11 Maths Chapter 10 - Straight Lines

Crucial Practice Problems for CBSE Class 11 Maths Chapter 10: Straight Lines

Important Questions for Class 11 Maths Chapter 10 - Straight Lines are available in Vedantu. These Questions are designed as per the latest Syllabus of NCERT Curriculum and are frequently asked in exams in different types. Solving these important questions will definitely help students to score good marks in final exams. All the solutions are explained in a detailed manner along with the graphical representation. Students can refer to these solutions for learning the most important questions and prepare for their board exams.

Download CBSE Class 11 Maths Important Questions 2024-25 PDF

Also, check CBSE Class 11 Maths Important Questions for other chapters:

Topics Covered in Class 11 Maths Chapter 10 - Straight Lines are as follows:

Slope of a Line

Slope of a line when coordinates of any two points on the line are given

Conditions for parallelism and perpendicularity of lines in terms of their slopes

Angle between two lines

Collinearity of three points

Various Forms of the Equation of a Line

Horizontal and vertical lines

Point-slope form

Two-point form

Slope-intercept form

Intercept – form

Normal form

General Equation of a Line

Different forms of Ax + By + C = 0f

Distance of a Point From a Line

Distance between two parallel lines

Study Important Questions for Class 11 Mathematics Chapter 10 - Straight Lines

Very Short Answer Questions. (1 Mark)

1. Find the slope of the lines passing through the point $\left( {3,\, - 2} \right)$ and $\left( { - 1,\,4} \right)$

Ans: Slope of the line = $m$

The given points are $\left( {3,\, - 2} \right)$ and $\left( { - 1,\,4} \right)$.

Let $\left( {{x_1},\,\,{y_1}} \right)$ be $\left( {3,\, - 2} \right)$ and $\left( {{x_2},\,\,{y_2}} \right)$ be $\left( { - 1,\,4} \right)$.

$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

$\Rightarrow \dfrac{{4 - \left( { - 2} \right)}}{{ - 1 - 3}}$

$\Rightarrow \dfrac{6}{{ - 4}} =  - \dfrac{3}{2}$  

Therefore, the slope of the line passing through the point $\left( {3,\, - 2} \right)$ and $\left( { - 1,\,4} \right)$ is $ - \dfrac{3}{2}$.

2. Three points $P\left( {h,\,\,k} \right),\,\,Q\left( {{x_1},\,\,{y_1}} \right)$ and $R\left( {{x_2},\,\,{y_2}} \right)$ lie on a line. Show that 

$\left( {h - {x_1}} \right)\left( {{y_2} - {y_1}} \right) = \left( {k - {y_1}} \right)\left( {{x_2} - {x_1}} \right)$

Ans: The given points are $P\left( {h,\,\,k} \right),\,\,Q\left( {{x_1},\,\,{y_1}} \right)$and $R\left( {{x_2},\,\,{y_2}} \right)$.

Since, the points $P,\,\,Q$ and $R$ lie on a line. Therefore, they are collinear.

Slope of $PQ$= Slope of $QR$

$ \Rightarrow \dfrac{{{y_1} - k}}{{{x_1} - h}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$                           $\left[ {\because \,\,{\text{Slope }} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right]$

$ \Rightarrow \dfrac{{\left( {k - {y_1}} \right)}}{{\left( {h - {x_1}} \right)}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$                        (On cross multiplication)

3. Write the equation of the line through the points $\left( {1, - 1} \right)$ and \[\left( {3,\,\,5} \right)\]

Ans: We know that equation of line through two points $\left(x_{1}, y_{1}\right) \&\left(x_{2}, y_{2}\right)$ is

$y-y_{1}=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$

Since the equation of the line through the points $(1,-1)$ and $(3,5)$.

$x_{1}=1, y_{1}=-1$

$x_{2}=3 \text { and } y_{2}=5$

Putting values

$(y-(-1))=\dfrac{5-(-1)}{3-1}(x-1)$

$y+1=\dfrac{5+1}{2}(x-1)$

$y+1=\dfrac{6}{2}(x-1)$

$y+1=3(x-1)$

$y+1=3 x-3$

$y-3 x+1+3=0$

$y-3 x+4=0$

Hence the required equation is $\mathrm{y}-3 \mathrm{x}+4=0$

4. Find the measure of the angle between the lines $x + y + 7 = 0$ and $x - y + 1 = 0$.

Ans: The given equation of lines are $x + y + 7 = 0$ and $x - y + 1 = 0$.

Express the given equation as slope-intercept form $y = mx + c$

(Slope)$m$= coefficient of $x$

$\mathrm{x}+\mathrm{y}+7=0$

$\mathrm{~m}_{1}=\dfrac{-1}{1}$

$\mathrm{x}-\mathrm{y}+1=0$

$\mathrm{~m}_{2}=\dfrac{-1}{-1}=1$

Product of these two slopes is $-1$, therefore, the lines are at right angles.

5. Find the equation of the line that has $y - $intercept $4$ and is perpendicular to the line $y = 3x - 2$.

Ans: The given equation of the is $y = 3x - 2$.

${m_1} = 3$

When the lines are perpendicular. Then the product of slope is $ - 1$.

$\therefore \,\,\,{m_1}.{m_2} =  - 1$

$\,\,\,\,\,\,\,\,\,\,3.{m_2} =  - 1$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{m_2} =  - \dfrac{1}{3}$

$y - $intercept of other line is 4.

Therefore, the required equation of the line using the slope-intercept form $y = mx + c$.

$y =  - \dfrac{1}{3}x + 4$

6. Find the equation of the line, which makes intercepts $ - 3$ and 2 on the $x$ and $y$-axis respectively.

Ans: Given,

($x - $intercept) a $ =  - 3$

($y - $intercept) b $ = 2$

Required equation is given by $\dfrac{x}{a} + \dfrac{y}{b} = 1$

$a =  - 3,b = 2$

$\therefore \dfrac{x}{{ - 3}} + \dfrac{y}{2} = 1$

$2x - 3y + 6 = 0$

7. Equation of a line is $3x - 4y + 10 = 0$ find its slope.

Equation of the line is $3x - 4y + 10 = 0$.

Slope of the line is given by $m = \dfrac{{ - \,{\text{co - efficient of }}x}}{{{\text{co - efficient of }}y}}$

$ \Rightarrow \dfrac{{ - 3}}{{ - 4}} = \dfrac{3}{4}$

8. Find the distance between the parallel lines $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$.

Equations of the parallel lines are $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$.

The general equation of the parallel lines is given by $Ax + By + {C_1} = 0{\text{ and }}Ax + By + {C_2} = 0$.

On comparing the given equation of two parallel lines with general equation, we get

$A = 3,\,\,B =  - 4,\,{C_1} = 7\,{\text{and }}{C_2} = 5$. 

Distance between two parallel lines is $d = \dfrac{{\left| {{C_1} - {C_2}} \right|}}{{\sqrt {{A^2} + {B^2}} }}$

$\Rightarrow \dfrac{{|7 - 5|}}{{\sqrt {{{(3)}^2} + {{( - 4)}^2}} }}$

$\Rightarrow \dfrac{2}{{\sqrt {9 + 16} }}$

$\Rightarrow \dfrac{2}{5}$

9. Find the equation of a straight line parallel to the $y$-axis and passing through the point $\left( {4,\, - 2} \right)$.

Ans: Given point is $\left( {4,\, - 2} \right)$

Equation of line parallel to $y$-axis is $x = a\,\,\,\,\,...(i)$

Since, equation (i) passing through $\left( {4,\, - 2} \right)$

$\therefore \,\,a = 4$

$x - 4 = 0$

10. If $3x - by + 2 = 0$ and $9x + 3y + a = 0$ represent the same straight line, find the values of $a$ and $b$.

Ans: Given the equations of the line are $3x - by + 2 = 0$ and $9x + 3y + a = 0$. 

According to the question,

$\dfrac{3}{9} = \dfrac{{ - b}}{3} = \dfrac{2}{a}$

On comparing the values, we get

$\dfrac{3}{9} = \dfrac{{ - b}}{3}\,\,\,\,\,{\text{and }}\dfrac{3}{9} = \dfrac{2}{a}$

$\therefore \,\,b =  - 1\,\,{\text{and }}\,a = 6{\text{ }}$

11. Find the distance between $P\left( {{x_1}{y_1}} \right)$ and $Q\left( {{x_2},{y_2}} \right)$ when ${\mathbf{PQ}}$ is parallel to the $y$-axis.

Ans. When PQ is parallel to the $y$-axis,

Then ${x_1} = {x_2}$

$\therefore PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $

$= \sqrt {{{\left( {{x_2} - {x_2}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$

$= \left| {{y_2} - {y_1}} \right|$

12. Find the slope of the line, which makes an angle of ${30^\circ }$ with the positive direction of $y - $axis measured anticlockwise.

Line which makes an angle of ${30^\circ }$ with the positive direction of $y - $axis measured anticlockwise.

Let $\theta $ be the inclination of the line,

$\theta  = {120^\circ }$

(slope) $m = \tan 120^\circ $

$= \tan \left( {90 + 30} \right)$

$=  - \sqrt 3$

13. Determine $x$ so that the inclination of the line containing the points $(x, - 3)$ and $(2,5)$ is ${135^\circ }$.

Ans: We have been given a line joining the points $(x,-3)$ and $(2,5)$ which makes an angle of $135^{\circ}$ with the $x$-axis.

We know that the slope of a line joining the two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is equal to the tangent of the angle made by the line with $x$-axis in anticlockwise direction given by as follows: slope $=\tan \theta=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

So we have $\theta=135^{\circ}, x_{1}=2, x_{2}=x, y_{1}=5, y_{2}=-3$

$\Rightarrow \tan 135^{\circ}=\dfrac{-3-5}{x-2}$

Since we know that $\tan 135^{\circ}=\tan \left(90^{\circ}+45^{\circ}\right)=-\cot 45^{\circ}=-1$ as in second quadrant tangent function is negative.

$\Rightarrow-1=\dfrac{-3-5}{x-2}$

$\Rightarrow-1=\dfrac{-8}{x-2}$

On cross multiplication, we get as follows:

$\Rightarrow-x+2=-8$

On adding $-2$ to both the sides of the equation, we get as follows:

$\Rightarrow-x-2+2=-8-2$

$\Rightarrow-x=-10$

$\Rightarrow x=10$

Therefore the value of $\mathrm{x}$ is equal to 10 .

14. Find the distance of the point $(4,1)$ from the line $3x - 4y - 9 = 0$

Point $(4,1)$ and equation of the line $3x - 4y - 9 = 0$

Let ${\text{d}}$ be the required distance,

$d = \dfrac{{|3 \cdot (4) - 4(1) - 9|}}{{\sqrt {{{(3)}^2} + {{( - 4)}^2}} }}$

$ = \dfrac{{| - 1|}}{5} = \dfrac{1}{5}$

15. Find the value of $x$ for which the points $(x, - 1),(2,1)$ and $(4,5)$ are collinear.

Ans. Given three collinear points.

Let the point be $A(x, - 1),B(2,1),C(4,5)$

Since, the points are collinear. Therefore,

Slope of ${\text{AB}} = $ Slope of ${\text{BC}}$

$\dfrac{{1 + 1}}{{2 - x}} = \dfrac{{5 - 1}}{{4 - 2}}$

$\dfrac{2}{{2 - x}} = \dfrac{{{4}}}{2}$

$2 - x = 1$

$ - x =  - 1$

16. Find the angle between the $x$-axis and the line joining the points $(3, - 1)$ and $\left( {4,\, - 2} \right)$.

Line joining the points $(3, - 1)$ and $\left( {4,\, - 2} \right)$.

${m_1} = 0\quad $ Slope of $x$-axis

${m_2} = $ slope of line joining points $(3, - 1)$ and $(4, - 2)$

${m_2} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

$ = \dfrac{{ - 2 - ( - 1)}}{{4 - 3}} =  - 1$

Angle between two line is given by $\tan \theta  = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$

$\tan \theta  = \left| {\dfrac{{-1 + 0}}{{1 + 0 \times ( - 1)}}} \right|$

$\tan \theta  = 1$

$\theta  = {45^\circ }$

17. Using slopes, find the value of $x$ for which the points $(x, - 1),\,(2,1)$ and $(4,5)$ are collinear.

Three collinear points.

Let the points be A$\left( {x,\, - 1} \right)$, B$\left( {2,\,\,1} \right)$ and C $\left( {4,\,\,5} \right)$.

$\Rightarrow \dfrac{{1 - \left( { - 1} \right)}}{{2 - x}} = \dfrac{{5 - 1}}{{4 - 2}}$

$\Rightarrow \,\,\,\,\,\,\,\,\,\dfrac{{1 + 1}}{{2 - x}} = \dfrac{4}{2} $

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 = 2\left( {2 - x} \right) $

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 = 2 - x $

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1$

18. Find the value of $K$ so that the line $2x + ky - 9 = 0$ may be parallel to $3x - 4y + 7 = 0$

The Equations of two parallel lines are $2x + ky - 9 = 0$ and $3x - 4y + 7 = 0$.

Slope of 1st line = slope of 2nd line

$\dfrac{{ - 2}}{k} = \dfrac{{ - 3}}{{ - 4}}$

$ \Rightarrow k = \dfrac{{ - 8}}{3}$

19. Find the value of $K$, given that the distance of the point $(4,1)$ from the line $3x - 4y + K = 0$ is 4 units.

 The distance of the point $(4,1)$ from the line $3x - 4y + k = 0$ is 4 units.

$\Rightarrow \dfrac{{|3(4) - 4(1) + k|}}{{\sqrt {{{(3)}^2} + {{( - 4)}^2}} }} = 4 $

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\dfrac{{\left| {12 - 4 + k} \right|}}{{\sqrt {25} }} = 4 $

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{{\left| {8 + k} \right|}}{5} = 4 $

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| {8 + k} \right| = 20$

$8 + k = 20$

  $\,\,\,\,\,\,\,k = 12$

$- \left( {8 + k} \right) = 20$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k =  - 28$

20. Find the equation of the line through the intersection of $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ which cuts off equal intercepts on the axes.

Equation of two intersecting lines $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$.

Slope of a line which makes equal intercept on the axes is $ - 1$ any line through the intersection of given lines is,

$(3x - 4y + 1) + K(5x - y - 1) = 0$

$(3 + 5K)x + y(K - 4) + 1 - K = 0$

$\Rightarrow \,\,\,\,m =  - \dfrac{{(3 + 5K)}}{{K - 4}}$

$\Rightarrow  - 1 =  - \dfrac{{(3 + 5K)}}{{K - 4}}$

$\Rightarrow \,\,\,K = \dfrac{{ - 7}}{4}$

21. Find the distance of the point $(-2,\,\,3)$ from the line $12x - 5y = 2$.

Ans: Given point $(-2,\,\,3)$ and equation of the line $12x - 5y = 2$.

Since, we know

Distance of a point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ from the line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ is $\dfrac{\left|\mathrm{ax}_{1}+\mathrm{by}_{1}+\mathrm{c}\right|}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}$ Distance of the point $(-2,3)$ from the line $12 \mathrm{x}-5 \mathrm{y}-2=0$ is

$\dfrac{|12(-2)-5(3)-2|}{\sqrt{12^{2}+(-5)^{2}}}=\dfrac{|-24-15-2|}{13}=\dfrac{41}{13}$

22. Find the equation of a line whose perpendicular distance from the origin is 5 units and angle between the positive direction of the $x$-axis and the perpendicular is ${30^\circ }$.

Ans. Given,

Perpendicular distance of the line from the origin is 5 units.

Angle between the positive direction of the $x$-axis and the perpendicular is ${30^\circ }$.

$p = 5,\alpha  = {30^\circ }$

Required equation is given by $x\cos \alpha  + y\sin \alpha  = p$.

$x\cos {30^\circ } + y\sin {30^\circ } = 5$

$\sqrt 3 x + y - 10 = 0$

23. Write the equation of the lines for which $\tan \theta  = \dfrac{1}{2}$, where ${\mathbf{Q}}$ is the inclination of the line and $x$ intercept is 4.

Slope of the line will become;

$m = \tan \theta  = \dfrac{1}{2}$ and 

$x - $intercept, i.e., d = 4

$y = \dfrac{1}{2}(x - 4)\,\,\,\,\,\,\,\,\,\left[ {\,\because y = m(x - d)} \right]$

$2y - x + 4 = 0$

24. Find the angle between the $x$-axis and the line joining the points $(3, - 1)$ and $(4, - 2)$.

Two distinct points.

Let $A(3, - 1)$ and $B(4, - 2)$

Slope of AB,

 $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

$\,\,\,\,\, = \dfrac{{ - 2 - ( - 1)}}{{4 - 3}}$

$\,\,\,\,\, =  - 1$

$\,\,\,\,\,\,\because m = \tan \theta$

$\therefore \tan \theta  =  - 1$

$\theta=135$

where  $\theta $ is the angle which AB makes with the positive direction of the x axis.

25. Find the equation of the line intersecting the $x$-axis at a distance of 3 units to the left of origin with slope $ - 2$.

Line intersecting the $x$-axis at a distance of 3 units to the left of origin.

Therefore, the point is $\left( { - 3,\,0} \right)$

Slope of the line = m $ =  - 2$

Required equation is given by one-point form,

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y - {y_1} = m\left( {x - {x_1}} \right)$

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,y - 0 =  - 2(x + 3)$

$\Rightarrow 2x + y + 6 = 0$

Short Answer Question (4 Marks)

1. If $p$ is the length of the $ \bot $ from the origin on the line whose intercepts on the axes are a and b . Show that $\dfrac{1}{{{p^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}$

Ans: Equation of the line is $\dfrac{x}{a} + \dfrac{y}{b} = 1$

$ \Rightarrow \dfrac{x}{a} + \dfrac{y}{b} - 1 = 0$

The distance of this line from the origin is P

$\therefore P = \dfrac{{\left| {\dfrac{0}{a} + \dfrac{0}{b} - 1} \right|}}{{\sqrt {{{\left( {\dfrac{1}{a}} \right)}^2} + {{\left( {\dfrac{1}{b}} \right)}^2}} }}\quad \left[ {d = \dfrac{{|ax + by + c|}}{{\sqrt {{a^2} + {b^2}} }}} \right]$

$\dfrac{P}{1} = \dfrac{1}{{\sqrt {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} }}$

$\dfrac{1}{P} = \sqrt {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} $

Squaring both the sides, 

$\dfrac{1}{{{P^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}$

Hence proved.

2. Find the value of $p$ so that the three lines $3x + y - 2 = 0,px + 2y - 3 = 0$ and $2x - y - 3 = 0$ may intersect at one point.

Equation of three lines,

$3x + y - 2 = 0\,\,\,\,\,\,\,.....(i)$

$px + 2y - 3 = 0\,\,\,\,\,.....(ii)$

$2x - y + 3 = 0\,\,\,\,\,\,\,\,.....(iii)$

On solving equation (i) and (iii), 

$x = 1$ and $y =  - 1$

Put x, y in equation (ii), 

$\,\,\,\,\,\,\,\,\,\,px + 2y - 3 = 0$

$p.1 + 2.\left( { - 1} \right) - 3 = 0$

$p - 2 - 3 = 0$

$\,\,\,\,\,\,\,\,p - 5 = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p = 5$

3. Find the equation to the straight line which passes through the point $(3,4)$ and has an intercept on the axes equal in magnitude but opposite in sign.

Ans: Given the makes equal intercepts in axes but opposite in sign.

Let the intercepts be,

($x - $intercept) a $ = a$

($y - $intercept) b $ = b$

The equation of the line is given by $\dfrac{x}{a} + \dfrac{y}{b} = 1$.

On substituting the values,

$\dfrac{x}{a} + \dfrac{y}{{ - a}} = 1$

$ \Rightarrow x - y = a \ldots  \ldots .(i)$

Since it passes through the point $(3,4)$

$3 - 4 = a$

$a =  - 1$

Put the value of a in equation (i)

$x - y =  - 1$

$x - y + 1 = 0$

4. By using the area of $\Delta .$ Show that the points $(a,b + c),\,\,(b,c + a)$ and $(c,a + b)$ are collinear.

Three points, let the points be ${\text{A}}(a,b + c),\,\,{\text{B}}(b,c + a)$ and ${\text{C}}(c,a + b)$

We know that

Area of triangle is given by 

$\Delta=\dfrac{1}{2} \begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{vmatrix}$

$x_{1}=a, y_{1}=b+c$

$x_{2}=b, y_{2}=c+a$

$x_{3}=c, y_{3}=a+b$

$\Delta=\dfrac{1}{2} \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1\end{vmatrix}$

Applying $C_{1} \rightarrow C_{1}+C_{2}$

$\Delta=\dfrac{1}{2} \begin{vmatrix} a+b+c & b+c & 1 \\ b+c+a & c+a & 1 \\ c+a+b & a+b & 1\end{vmatrix}$

Taking $(a+b+c)$ common from $C_{1}$

$\Delta=\dfrac{1}{2}(a+b+c) \begin{vmatrix} 1 & b+c & 1 \\ 1 & c+a & 1 \\ 1 & a+b & 1\end{vmatrix}$

Here, $1^{\text {st }}$ and $3^{\text {rd }}$ Column are Identical

Hence value of determinant is zero

$\Delta=\dfrac{1}{2}(a+b+c) \times 0 $

So, $\Delta=\mathbf{0}$

Hence points $A, B \& C$ are collinear

5. Find the slope of a line, which passes through the origin, and the midpoint of the line segment joining the point $P(0, - 4)$ and $Q(8,0)$

Line segment joining the point $P(0, - 4)$ and $Q(8,0)$.

Let $M$ be the midpoint of segment PQ then

$M = \left( {\dfrac{{{x_1} + {x_2}}}{2} + \dfrac{{{y_1} + {y_2}}}{2}} \right)$

$\,\,\,\,\,\,\,\,\,\, = \left( {\dfrac{{0 + 8}}{2},\dfrac{{ - 4 + 0}}{2}} \right)$

$\,\,\,\,\,\,\,\,\,\,\, = \left( {4, - 2} \right)$

Slope of line joining origin $O\left( {0,\,0} \right)$ and the mid-point $M\left( {4, - 2} \right)$,

 $OM = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

$\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{ - 2 - 0}}{{4 - 0}}$

$\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{ - 1}}{2}$

6. Find the equation of the line passing through the point $(2,2)$ and cutting off intercepts on the axes whose sum is 9.

Ans: Required equation is  $\dfrac{x}{a} + \dfrac{y}{b} = 1 \ldots  \ldots (i)$

($x - $intercept) $ = a$

($y - $intercept) $ = b$

Sum of intercepts = 9

$\therefore \,\,a + b = 9$

$b = 9 - a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( {ii} \right)$

Substituting the value of equation (ii) in equation (i), 

$ \Rightarrow \dfrac{x}{a} + \dfrac{y}{{9 - a}} = 1$

This line passes through $(2,2)$

$\therefore \dfrac{2}{a} + \dfrac{2}{{9 - a}} = 1$

${a^2} - 9a + 18 = 0$

${a^2} - 6a - 3a + 18 = 0$

$a(a - 6) - 3(a - 6) = 0$

$(a - 6)(a - 3) = 0$

$a = 6\,\,\,{\text{or}}\,\,\,a = 3$

On substituting the values of a in equation (ii), we get

$b = 3\,\,\,\,{\text{or }}\,\,b = 6$

Therefore, the equation of line are,

$\dfrac{x}{6} + \dfrac{y}{3} = 1\quad \dfrac{x}{3} + \dfrac{y}{6} = 1$

7. Reduce the equation $\sqrt 3 x + y - 8 = 0$ into normal form. Find the values $p$ and $\omega $.

Equation of a line$\sqrt 3 x + y - 8 = 0$

Let $\sqrt 3 x + y = 8 \ldots  \ldots .(i)$

$\sqrt {{{(\sqrt 3 )}^2} + {{(1)}^2}}  = 2$

Dividing (i) by 2

$\dfrac{{\sqrt 3 }}{2}x + \dfrac{y}{2} = 4$

$x\cos {30^\circ } + y \cdot \sin {30^\circ } = 4 \ldots  \ldots .(ii)$

On comparing the above equation with the standard from,

$x\cos \omega  + y\sin \omega  = p$

$p$is the perpendicular distance from the origin

$\omega $ is the angle between perpendicular and the positive $x - $axis

$\omega  = \dfrac{\pi }{6}$

8. Without using the Pythagoras theorem show that the points $(4,4),(3,5)$ and $( - 1, - 1)$ are the vertices of a right angled $\Delta $.

Ans: Let given points be $A(4,4),B(3,5)$ and $C( - 1, - 1)$

${m_1} = \dfrac{{5 - 4}}{{3 - 4}} =  - 1$

Slope of BC,

${m_2} = \dfrac{{ - 1 - 5}}{{ - 1 - 3}}$

$\Rightarrow \dfrac{{ - 6}}{{ - 4}} = \dfrac{3}{2}$

Slope of AC,

${m_3} = \dfrac{{ - 1 - 4}}{{ - 1 - 4}} =  + 1$

Since, 

Slope of $AB \times $ slope of $AC =  - 1$

${m_1} \times {m_3} =  - 1$

$ \Rightarrow AB \bot AC$                (if the product of slope is $ - 1$, then the lines are perpendicular)

Hence $\Delta {\text{ABC}}$ is right angled at ${\text{A}}$.

9. The owner of a milk store finds that he can sell 980 litres of milk each week at Rs 14 per litre and 1220 litre of milk each week at Rs 16 per litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17 per litre?

Ans. Let selling price be P along $x$-axis

 demand of milk be D along $y$-axis

We know that the equation of line is

Here, $P$ is along $x$-axis and $D$ is along $y$-axis

So, our equation becomes

Owner sells 980 litre milk at Rs 14 /litre

So, $D=980 \& P=14$ satisfies the equation

Putting values in (1)

$980=14 m+c$

Owner sells 1220 litre milk at Rs $16 /$ litre

So, $D=1220 \& P=16$ satisfies the equation

$1220=16 m+c$

So, our equations are

$980-14 m=c$

Putting value of $\mathrm{c}$ in $(\mathrm{B})$

$1220=16 m+980-14 m$

$1220-980=16 m-14 m$

$\dfrac{240}{2}=2$

Putting $m=120$ in $(A)$

$980-14(120)=c$

$980-1680=c$

Putting value of $m \& c$ in (1)

D=120 P-700

Hence, the required equation is $\mathrm{D}=120 \mathrm{P}-700$

We need to find how many litres could he sell weekly at Rs $17 /$ litre i.e. we need to find $D$ when $P=17$

Putting $P=17$ in the equation

D=120(17)-700

Hence when price is Rs $17 /$ litre, 1340 litres of milk could be sold.

10. The line through the points $({\text{h}},3)$ and $(4,1)$ intersects the line $7x - 9y - 19 = 0$ a right angle. Find the value of $h$.

Ans: Given a line joining $({\text{h}},3)$ and $(4,1)$

Let $\left( {{x_1},\,{y_1}} \right){\text{ and }}\left( {{x_2},\,{y_2}} \right)$ be $({\text{h}},3)$ and $(4,1)$ respectively.

Find slope by two-point form,

$ {m_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} $

 $  \Rightarrow \dfrac{{1 - 3}}{{4 - h}} = \dfrac{{ - 2}}{{4 - h}}$

Given line is $7x - 9y - 19 = 0$

${m_2} = \dfrac{{ - 7}}{{ - 9}}$

The lines intersect at right angles, i.e., product of slope $ =  - 1$

$\,\,\therefore \,\,\,\,\,\,\,\,\,\,m{{\kern 1pt} _1} \times {m_2} =  - 1$

$\left( {\dfrac{{ - 2}}{{4 - h}}} \right) \times \left( {\dfrac{7}{9}} \right) =  - 1$

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,h = \dfrac{{22}}{9}$

11. Find the equations of the lines, which cut off intercepts on the axes whose sum and products are 1 and $ - 6$ respectively.

Ans: Let the intercepts made by the line on the axes be,

Sum of intercepts, i.e., $a + b = 1\,\,\,\,\,\,\,\,.....(i)$ 

Product of intercepts, i,e., $ab =  - 6\,\,\,\,\,\,\,\,.....\left( {ii} \right)$

$b = 1 - a\quad [$ from $(i)]$

Put b in equation (ii)

$a(1 - a) =  - 6$

$a - {a^2} =  - 6$

${a^2} - a - 6 = 0$

$(a - 3)(a + 2) = 0$

$a = 3, - 2$

When $a = 3$

$b =  - 2$

Required equation is  $\dfrac{x}{a} + \dfrac{y}{b} = 1$

$\dfrac{x}{3} + \dfrac{y}{{ - 2}} = 1$

$2x - 3y - 6 = 0$

When $a =  - 2$

$\dfrac{x}{{ - 2}} + \dfrac{y}{3} = 1$

$3x - 2y + 6 = 0$

12. The slope of a line is double of the slope of another line. If the tangent of the angle between them is $\dfrac{1}{3}$, find the slopes of the lines.

The slope of a line is double of the slope of another line.

Tangent of the angle between them, i.e., $\tan \theta  = \dfrac{1}{3}$

Let the slope of one line is $m$ and other line is 2m

$\dfrac{1}{3} = \left| {\dfrac{{2m - m}}{{1 + (2m)(m)}}} \right|$

$\dfrac{1}{3} = \left| {\dfrac{m}{{1 + 2{m^2}}}} \right|$

$ \pm \dfrac{1}{3} = \dfrac{m}{{1 + 2{m^2}}}$

$\dfrac{1}{3} = \dfrac{m}{{1 + 2{m^2}}}$

$2{m^2} - 3m + 1 = 0$

$2{m^2} - 2m - m + 1 = 0$

$2m(m - 1) - 1(m - 1) = 0$

$(m - 1)(2m - 1) = 0$

$m = 1,\,\,m = \dfrac{1}{2}$

$\dfrac{{ - 1}}{3} = \dfrac{m}{{1 + 2{m^2}}}$

$ - 1 - 2{m^2} = 3m$

$2{m^2} + 3m + 1 = 0$

$2{m^2} + 2m + m + 1 = 0$

$2m(m + 1) + 1(m + 1) = 0$

$(m + 1)(2m + 1) = 0$

$m =  - 1,\,\,m = \dfrac{{ - 1}}{2}$

13. Point $R(h,k)$ divides a line segment between the axes in the ratio $1:2$. Find the equation of the line.

Ans: It is given that $R(h,k)$ divides ${\text{AB}}$ in the ratio $1:2$

Required equation is  $\dfrac{x}{a} + \dfrac{y}{b} = 1\,\,\,\,\,.....\left( i \right)$

$\therefore \left( {h,\,k} \right) = \left( {\dfrac{{2a}}{3},\,\dfrac{b}{3}} \right) $

$h = \dfrac{{2a}}{3}\,\,\,\,{\text{and }}\,\,\,k = \dfrac{b}{3} $

$\dfrac{{3h}}{2} = a\,\,\,{\text{and}}\,\,\,3k = b$

Put $a$ and $b$ in equation   ……(i)

$\dfrac{x}{{\dfrac{{3h}}{2}}} + \dfrac{y}{{3k}} = 1 $

$\dfrac{{2x}}{h} + \dfrac{y}{k} = 3$

14. The Fahrenheit temperature F and absolute temperature K satisfy a linear equation. Given that ${\text{K}} = 273$ when ${\text{F}} = 32$ and that ${\text{K}} = 373$ when ${\text{F}} = 212$. Express ${\text{K}}$ in terms of ${\text{F}}$ and find the value of ${\text{F}}$ when ${\text{K}} = 0$

Ans: Let Fahrenheit temperature ${\text{F}}$ along $x$-axis and absolute temperature ${\text{K}}$ along $y$-axis.

Let $\left( {{x_1},\,{y_1}} \right){\text{ and }}\left( {{x_2},\,{y_2}} \right)$ be $\left( {273,\,32} \right)$ and $\left( {373,\,212} \right)$ respectively.

Using equation of straight line by two-point form,

$y - {y_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)$

$K - 273 = \dfrac{{373 - 273}}{{212 - 32}}(F - 32)$

$K - 273 = \dfrac{{10}}{{18}}(F - 32)$

When $K = 0$, then

$\,\,\,\,K = \dfrac{5}{9}(F - 32) + 273 $

$\therefore F = \dfrac{9}{5}\left( {K - 273} \right) + 32\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \,\,K = 0} \right] $

$\,\,\,\,\,\,\,\,\, = \dfrac{9}{5} \times \left( { - 273} \right) + 32 $

$\,\,\,\,\,\,\,\,\, =  - 491.4 + 32 $

$\,\,\,\,\,\,\,\,\, =  - 459.4$

15. If three points $(h,0),\,\,(a,b)$ and $(0,k)$ lie on a line, show that $\dfrac{a}{h} + \dfrac{b}{k} = 1$

Ans: Let $A(h,0),\,\,B(a,b)$ and ${\text{C}}(0,k)$

Since, the three points lie on the same line. Therefore,

Slope of ${\text{AB}} = $ slope of ${\text{BC}}$

$\Rightarrow \dfrac{b}{a-h}=\dfrac{k-b}{-a} $

$\Rightarrow (b)(-a)=(k-b)(a-h) $

$\Rightarrow(-a)(b)=ka-kh-ab+bh $

$\Rightarrow ka+bh=k h$

$\dfrac{{ak}}{{hk}} + \dfrac{{hb}}{{hk}} = 1$

$\dfrac{a}{h} + \dfrac{b}{k} = 1$

16. $P\left( {a,\,\,b} \right)$ is the midpoint of a line segment between axes. Show that the equation of the line is $\dfrac{x}{a} + \dfrac{y}{b} = 2$.

Mid-point of a line segment between axes $P\left( {a,\,\,b} \right)$.

Let P be the mid-point of $A\left( {c,\,0} \right)$ and $B\left( {0,\,d} \right)$

Required equation is  $\dfrac{x}{a} + \dfrac{y}{b} = 1$ i.e., $\dfrac{x}{c} + \dfrac{y}{d} = 1$.

$x - $intercept = a

$y - $intercept = b

Coordinate of $P = \left( {\dfrac{c}{2},\dfrac{d}{2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because {\text{ mid - point}} = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)} \right]$

$\therefore \,\,(a,b) = \left( {\dfrac{c}{2},\dfrac{d}{2}} \right)$

$\dfrac{a}{1} = \dfrac{c}{2}$

$\dfrac{b}{1} = \dfrac{d}{2}$

Put the value of $c$ and $d$ in equation (i)

$\dfrac{x}{{2a}} + \dfrac{y}{{2b}} = 1$

$\,\,\,\,\,\dfrac{x}{a} + \dfrac{y}{b} = 2$

17. The line $ \bot $ to the line segment joining the points $(1,0)$ and $(2,3)$ divides it in the ratio $1:n$ find the equation of the line.

Coordinates of line segments.

Let $A\left( {1,\,\,0} \right)$ and $B\left( {2,\,\,3} \right)$. Ratio = $1:n$

Coordinate of $C\left( {\dfrac{{2 + n}}{{1 + n}} = \dfrac{3}{{1 + n}}} \right)$

${m_{AB}} = 3$

Slope of PQ,

${m_{PQ}} =  - \dfrac{1}{3}$

Equation of ${\text{PQ}}$using one-point form,

$\left( {{x_1},\,{y_1}} \right) = \left( {\dfrac{{2 + n}}{{1 + n}} = \dfrac{3}{{1 + n}}} \right)$

$\,\,\,\,\,\,\,{m_{PQ}} =  - \dfrac{1}{3}$

$y - {y_1} = m\left( {x - {x_1}} \right)$

$\dfrac{y}{1} - \dfrac{3}{{1 + n}} =  - \dfrac{1}{3}\left( {\dfrac{x}{1} - \dfrac{{2 + n}}{{1 + n}}} \right)$

$(n + 1)x + 3(n + 1)y - (n + 11) = 0$

Long Answer Question (6 Marks) 

1. Find the values of $k$ for the line $(k - 3)x - \left( {4 - {k^2}} \right)y + {k^2} - 7k + 6 = 0$

(a). Parallel to the $x$-axis

(b). Parallel to $y$-axis

(c). Passing through the origin

Given, equation of line $(k - 3)x - \left( {4 - {k^2}} \right)y + {k^2} - 7k + 6 = 0$

(a) The line is parallel to $x$-axis, if coefficient of $x = 0$

coefficient of $x = k - 3$

$\therefore \,\,k - 3 = 0$

$\,\,\,\,\,\,\,\,\,\,\,k = 3$

(b) The line is parallel to $y$-axis, if coefficient of $y = 0$

coefficient of $y = 4 - {k^2}$

$\therefore \,\,4 - {k^2} = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,k =  \pm 2$

(c) line passes through the origin if $(0,0)$ lies on given equation,

$(k - 3) \cdot (0) - \left( {4 - {k^2}} \right)(0) + {k^2} - 7k + 6 = 0$

$(k - 6)(k - 1) = 0$

2. If $p$ and $q$ are the lengths of $ \bot $ from the origin to the lines $x\cos \theta  - y\sin \theta  = k\cos 2\theta $ and $x\sec \theta  + y\operatorname{cosec} \theta  = k$ respectively, prove that ${p^2} + 4{q^2} = {k^2}$.

Equation of lines are,

$x \cos \theta-y \sin \theta=k \cos 2 \theta \ldots . .(1)$

$\mathrm{x} \sec \theta+\mathrm{y} \csc \theta=\mathrm{k} \ldots .(2)$

The general equation is of the from ${\text{Ax}} + {\text{By}} + {\text{C}} = 0$.

The perpendicular distance $d$ of a line from a point $\left( {{{\text{x}}_1},{{\text{y}}_1}} \right)$ is given by,

${\text{d}} = $ $\dfrac{{\left| {{\text{A}}{{\text{x}}_1} + {\text{B}}{{\text{y}}_1} + {\text{C}}} \right|}}{{\sqrt {{{\text{A}}^2} + {{\text{B}}^2}} }}$

On comparing equation (1) to the general equation of line i.e.,

 $Ax + By + C = 0$, we obtain 

$A = \cos \theta ,\,\,{\text{B}} =  - \sin \theta $, and ${\text{C}} =  - {\text{k}}\cos 2\theta $

It is given that ${\text{p}}$ is the length of the perpendicular from $\left( {0,\,\,0} \right)$ to line (1) 

$\therefore {\text{p}} = \dfrac{{|{\text{A}}(0) + {\text{B}}(0) + {\text{C}}|}}{{\sqrt {{{\text{A}}^2} + {{\text{B}}^2}} }}$

 $\Rightarrow \dfrac{{|{\text{C}}|}}{{\sqrt {{{\text{A}}^2} + {{\text{B}}^2}} }} = \dfrac{{| - {\text{k}}\cos 2\theta |}}{{\sqrt {{{\cos }^2}\theta  + {{\sin }^2}\theta } }}$

 $\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,| - {\text{k}}\cos 2\theta |\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots  \ldots ..(3)$

On comparing equation(2) to the general equation of line i.e.,

${\text{Ax}} + {\text{By}} + {\text{C}} = 0$, we obtain

$A = \sec \theta ,B = {\text{cosec}}\theta $, and $C =  - {\text{k}}$

It is given that ${\text{q}}$ is the length of the perpendicular from $(0,0)$ to line (2) $\therefore {\text{q}} = \dfrac{{|{\text{A}}(0) + {\text{B}}(0) + {\text{C}}|}}{{\sqrt {{{\text{A}}^2} + {{\text{B}}^2}} }}$

$\Rightarrow \dfrac{{|{\text{C}}|}}{{\sqrt {{{\text{A}}^2} + {{\text{B}}^2}} }} = \dfrac{{| - {\text{k}}|}}{{\sqrt {{{\sec }^2}\theta  + {{\csc }^2}\theta } }}\,\,\,\,\, \ldots ..(4)$

From (3) and (4) we have,

${{\text{p}}^2} + 4{{\text{q}}^2} = {\left( {| - {\text{k}}\cos 2\theta |} \right)^2} + 4{\left\{ {\dfrac{{| - {\text{k}}|}}{{\sqrt {{{\sec }^2}\theta  + {{\csc }^2}\theta } }}} \right\}^2}$

$ = {{\text{k}}^2}{\cos ^2}2\theta  + \dfrac{{4{{\text{k}}^2}}}{{\left( {{{\sec }^2}\theta  + {{\csc }^2}\theta } \right)}}$

$ = {{\text{k}}^2}{\cos ^2}2\theta  + \dfrac{{4{{\text{k}}^2}}}{{\left\{ {\dfrac{1}{{{{\cos }^2}\theta }} + \dfrac{1}{{{{\sin }^2}\theta }}} \right\}}}$

$ = {{\text{k}}^2}{\cos ^2}2\theta  + \dfrac{{4{{\text{k}}^2}}}{{\left\{ {\dfrac{{{{\sin }^2} + {{\cos }^2}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta }}} \right\}}}$

$ = {{\text{k}}^2}{\cos ^2}2\theta  + 4{{\text{k}}^2}{\sin ^2}\theta {\cos ^2}\theta $

$ = {{\text{k}}^2}{\cos ^2}2\theta  + {{\text{k}}^2}{(2\sin \theta \cos \theta )^2}$

$ = {{\text{k}}^2}{\cos ^2}2\theta  + {{\text{k}}^2}{\sin ^2}2\theta $

$ = {{\text{k}}^2}\left( {{{\cos }^2}2\theta  + {{\sin }^2}2\theta } \right) = {{\text{k}}^2}\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \,\,{{\cos }^2}\theta  + {{\sin }^2}\theta  = 1} \right]$

Hence proved ${{\text{p}}^2} + 4{{\text{q}}^2} = {{\text{k}}^2}$.

3. Prove that the product of the $ \bot $ drawn from the points $\left( {\sqrt {{a^2} - {b^2}} ,0} \right)$ and $\left( { - \sqrt {{a^2} - {b^2}} ,0} \right)$ to the line $\dfrac{x}{a}\cos \theta  + \dfrac{y}{b}\sin \theta  = 1$ is ${b^2}$.

Ans: Given equation of line is $\dfrac{x}{a}\cos \theta  + \dfrac{y}{b}\sin \theta  = 1$

Let ${p_1}$ be the distance from $\left( {\sqrt {{a^2} - {b^2}} ,0} \right)$ to the given line,

${p_1} = \dfrac{{\mid \dfrac{{\sqrt {{a^2} - {b^2}} }}{a} \cdot \cos \theta  - 1}}{{\sqrt {{{\left( {\dfrac{{\cos \theta }}{a}} \right)}^2} + {{\left( {\dfrac{{\sin \theta }}{b}} \right)}^2}} }}\,\,\,\,\,\,\left[ {\because  \bot {\text{ from the points }}\sqrt {{a^2} - {b^2}} ,0} \right]$

${p_2}$ be the distance from $\left( { - \sqrt {{a^2} - {b^2}} ,0} \right)$ to given line,

${p_2} = \dfrac{{\left| { - \dfrac{{\sqrt {{a^2} - {b^2}} }}{a}\cos \theta  - 1} \right|}}{{\sqrt {{{\left( {\dfrac{{\cos \theta }}{a}} \right)}^2} + {{\left( {\dfrac{{\sin \theta }}{b}} \right)}^2}} }}$

Product of perpendicular lines, i.e., ${p_1}.{p_2}$

${p_1}{p_2} = \dfrac{{\left| {\left( {\dfrac{{\sqrt {{a^2} - {b^2}} }}{a}\cos \theta  - 1} \right)\left( { - \dfrac{{\sqrt {{a^2} - {b^2}} }}{a}\cos \theta  - 1} \right)} \right|}}{{\dfrac{{{{\cos }^2}\theta }}{{{a^2}}} + \dfrac{{{{\sin }^2}\theta }}{{{b^2}}}}}$

$ = \dfrac{{\left| {\left( {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} \right) \cdot {{\cos }^2}\theta  - 1} \right|}}{{\dfrac{{{b^2}{{\cos }^2}\theta  + {a^2}{{\sin }^2}\theta }}{{{a^2}{b^2}}}}}$

$ = \dfrac{{\left| {{a^2}{{\cos }^2}\theta  - {b^2}{{\cos }^2}\theta  - {a^2}} \right|{a^2}{b^2}}}{{{a^2}\left( {{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta } \right)}}$

$= \dfrac{{\left| { - \left( {{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta } \right)} \right|{b^2}}}{{{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta }}\quad \left[ {\because \,\,{a^2}{{\cos }^2}\theta  - {a^2} = {a^2}\left( {{{\cos }^2}\theta  - 1} \right)} \right]$

$= \dfrac{{\left( {{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta } \right){b^2}}}{{{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta }}$

Hence proved that the product of the $ \bot $ drawn from the points $\left( {\sqrt {{a^2} - {b^2}} ,0} \right)$ and $\left( { - \sqrt {{a^2} - {b^2}} ,0} \right)$ to the line $\dfrac{x}{a}\cos \theta  + \dfrac{y}{b}\sin \theta  = 1$ is ${b^2}$.

4. Find equation of the line mid way between the parallel lines $9x + 6y - 7 = 0$ and $3x + 2y + 6 = 0$

Ans: The given equations of parallel lines are

$9x + 6y - 7 = 0$

$3\left( {3x + 2y - \dfrac{7}{3}} \right) = 0$

$3x + 2y - \dfrac{7}{3} = 0 \ldots  \ldots (i)$

$3x + 2y + 6 = 0 \ldots ..(ii)$

Let the eq. of the line mid way between the parallel lines (i) and (ii) be

$3x + 2y + k = 0 \ldots  \ldots .(iii)$

Distance between (i) and (iii) = distance between (ii) and (iii)

Distance between two parallel line,

$d = \dfrac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {{a^2} + {b^2}} }}$

$\left| {\dfrac{{k + \dfrac{7}{3}}}{{\sqrt {9 + 4} }}} \right| = \left| {\dfrac{{k - 6}}{{\sqrt {9 + 4} }}} \right|\left[ {\because d = \dfrac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {{a^2} + {b^2}} }}} \right]$

$k + \dfrac{7}{3} = k - 6$

$k = \dfrac{{11}}{6}$

Substituting the value of $k$ in equation (iii)

Therefore, the required equation is $3x + 2y + \dfrac{{11}}{6} = 0$.

5. Assuming that straight lines work as the plane mirror for a point, find the image of the point $(1,2)$ in the line $x - 3y + 4 = 0$

Ans: Let $Q(h,k)$ is the image of the point $P(1,2)$ in the line $x - 3y + 4 = 0 \ldots  \ldots (i)$

Coordinate of midpoint of $PQ = \left( {\dfrac{{h + 1}}{2},\dfrac{{k + 2}}{2}} \right)$

This point will satisfy the eq. .......(i)

$\left( {\dfrac{{h + 1}}{2}} \right) - 3\left( {\dfrac{{k + 2}}{2}} \right) + 4 = 0$

$\dfrac{{h + 1}}{2} - \dfrac{{3k + 6}}{2} + 4 = 0$

$h - 3k =  - 3 \ldots  \ldots .(ii)$

Since, the object and the line are perpendicular. Therefore,

(Slope of line ${\text{PQ}}) \times $ (slope of line $x - 3y + 4 = 0) =  - 1$

$\left( {\dfrac{{k - 2}}{{h - 1}}} \right)\left( {\dfrac{{ - 1}}{{ - 3}}} \right) =  - 1$

$3h + k = 5 \ldots  \ldots .(iii)$

On solving (ii) and (iii)

$h = \dfrac{6}{5}$ and $k = \dfrac{7}{5}$

6. A person standing at the junction (crossing) of two straight paths represented by the equations $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$ wants to reach the path whose equation is $6x - 7y + 8 = 0$ in the least time. Find the equation of the path that he should follow.

Ans: The given equations of parallel lines are,

$2x - 3y - 4 = 0$....... (i)

$3x + 4y - 5 = 0 \ldots  \ldots (ii)$

Person wants to reach the path whose equation is,

$6x - 7y + 8 = 0 \ldots  \ldots (iii)$

On solving eq. (i) and (ii)

we get $\left( {\dfrac{{31}}{{17}},\dfrac{{ - 2}}{{17}}} \right)$

To reach the line (iii) in least time the man must move along the perpendicular from crossing point $\left( {\dfrac{{31}}{{17}},\dfrac{{ - 2}}{{17}}} \right)$ to (iii) line,

Slope of line(iii) is $\dfrac{6}{7}$

Therefore, the slope of the required path,

 Slope of the required path $ \times $Slope of line(iii) $ =  - 1$

Slope of the required path $ \times $$\dfrac{6}{7}$$ =  - 1$

Slope of the required path$ =  - \dfrac{7}{6}$

Therefore, the equation of the lie using one-point form,

$\,\,\,\,y - \left( { - \dfrac{2}{{17}}} \right) =  - \dfrac{7}{6}\left( {x - \dfrac{{31}}{{17}}} \right)$

$119x + 102y = 205$

7. A line is such that its segment between the lines $5x - y + 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1,5)$ obtains its equation.

Ans: $P\left( {{x_1},\,\,{y_1}} \right)$ lies on $5x - y + 4 = 0$

$5{x_1} - {y_1} + 4 = 0$

and $Q\left( {{x_2}{y_2}} \right)$ lies on $3x + 4y - 4 = 0$

$3{x_2} + 4{y_2} - 4 = 0$

On simplifying,

${y_1} = 5{x_1} + 4$

${y_2} = \dfrac{{4 - 3{x_2}}}{4}$

Since ${\text{R}}$ is the mid-point of PQ,

$\dfrac{{{x_1} + {x_2}}}{2} = 1,\dfrac{{{y_1} + {y_2}}}{2} = 5$

${x_1} + {x_2} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( i \right)$

${y_1} + {y_2} = 10\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( {ii} \right)$

On substituting the values of ${y_1}{\text{ and }}{y_2}$ from above in equation (ii),

$5{x_1} + 4 + \dfrac{{4 - 3{x_2}}}{4} = 10$

$20{x_1} + 16 + 4 - 3{x_2} = 40$

$20{x_1} - 3{x_2} = 20\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( {iii} \right)$

On solving equation (i) and (iii), we get

${x_1} = \dfrac{{26}}{{23}},{x_2} = \dfrac{{20}}{{23}}$

and ${y_1} = \dfrac{{222}}{{23}},{y_2} = \dfrac{8}{{23}}$

$P\left( {\dfrac{{26}}{{23}},\,\,\dfrac{{222}}{{23}}} \right){\text{ and }}Q\left( {\dfrac{{20}}{{23}},\,\dfrac{8}{{23}}} \right)$

Equation of line PQ using two-point form,

$y - \dfrac{{222}}{{23}} = \dfrac{{\dfrac{8}{{23}} - \dfrac{{222}}{{23}}}}{{\dfrac{{20}}{{23}} - \dfrac{{26}}{{23}}}}\left( {x - \dfrac{{26}}{{23}}} \right)$

$107x - 3y - 92 = 0$

8. Find the equations of the lines which pass through the point $(4,5)$ and make equal angles with the lines $5x - 12y + 6 = 0$ and $3x - 4y - 7 = 0$

Ans: Let the equation of line passing through the point $(4,5)$ be,

$y - 5 = m\left( {x - 4} \right)\,\,\,\,\,.....\left( i \right)$

Given an equation of two lines $5x - 12y + 6 = 0$ and $3x - 4y - 7 = 0$.

slope of line $5x - 12y + 6 = 0$ is ${m_1} = \dfrac{5}{{12}}$

slope of line $3x - 4y - 7 = 0$ is ${m_2} = \dfrac{3}{4}$

Since, the line makes equal angles. Therefore,

$\tan \theta  = \dfrac{{m - {m_1}}}{{1 + m.{m_1}}}{\text{  and }}\tan \theta  = \dfrac{{m - {m_2}}}{{1 - m.{m_2}}}$

$\dfrac{{m - \dfrac{5}{{12}}}}{{1 + m.\dfrac{5}{{12}}}} = \dfrac{{m - \dfrac{3}{4}}}{{1 + m.\dfrac{3}{4}}}$

$m = \dfrac{{ - 7}}{4}\;\;{\text{and }}m = \dfrac{4}{7}$

Putting the above values in equation (i),

Required equations are

$y - 5 = \dfrac{4}{7}(x - 4){\text{   and  }}y - 5 = \dfrac{{ - 7}}{4}(x - 4)$

$4x - 7y + 19 = 0\,\,\,\,{\text{and}}\,\,\,7x + 4y - 48 = 0$

Important Questions from Straight Lines (Short, Long & Practice)

Short answer type questions.

1. Find the slope of the lines passing through the point (3,-2) and (-1,4)

2. Write the equation of the line through the points (1,-1) and (3,5)

3. Find the measure of the angle between the lines x+y+7=0 and x-y+1=0

Long Answer Type Questions

1. Find the equation to the straight line which passes through the point (3,4) and has an intercept on the axes equal in magnitude but opposite in sign. 

2. Find the equations of the lines, which cut off intercepts on the axes whose sum and product are 1 and -6 respectively.

3. Find equation of the line passing through the point (2,2) and cutting off intercepts on the axes whose sum is 9

Practice Questions

1. Find the equation of a straight line parallel to y-axis and passing through the point (4,-2)

2. Find the slope of the line, which makes an angle of 30 o with the positive direction of y-axis measured anticlockwise.

3. Find the equation of the line intersecting the x-axis at a distance of 3 units to the left of origin with slope -2.

Key Features of Important Questions for Class 11 Maths Chapter 10 - Straight Lines

All the questions are curated as per examination point of view to help students score better.

Solutions are explained in a step by step manner for all questions.

All solutions are easy to understand and learn as they are clearly written by subject experts to match the curriculum.

These important questions help in developing a good conceptual foundation for students, which is important in the final stages of preparation for board and competitive exams.

These solutions are absolutely free and available in a PDF format.

Important Related Links for CBSE Class 11 

Vedantu's Important Questions for Class 11 Maths Chapter 10 - Straight Lines prove to be an invaluable resource for students seeking comprehensive preparation. By meticulously curating essential concepts and problem-solving techniques, Vedantu empowers learners to master this fundamental topic with confidence. The well-structured questions encourage critical thinking and enhance problem-solving skills, enabling students to grasp the intricacies of straight lines effortlessly. Additionally, the platform's interactive approach fosters a deeper understanding of mathematical principles, promoting a strong foundation for future studies. Vedantu's commitment to academic excellence and accessibility makes it an ideal companion for students aspiring to excel in mathematics. With Vedantu's guidance, navigating the complexities of straight lines becomes an enriching and rewarding experience.

FAQs on Important Questions for Class 11 Maths Chapter 10 - Straight Lines

1. What is the significance of Vedantu’s extra questions for Class 11 Maths Chapter 10 Straight Lines?

Vedantu’s extra questions for CBSE Class 11 Maths Chapter 10 Straight Lines are helpful during the exam preparation. These important questions from Class 11 Maths Chapter 10 allow students to read the chapter thoroughly. Working on these questions will make students familiar with all types of questions that are important from an examination point of view. This PDF covers all the important topics of the chapters. These questions are based on the exam pattern and are added after referring to previous year question papers. The free PDF of CBSE Maths Class 11 Chapter 10 can be used at the time of revision and they are extremely helpful in scoring well in the paper.

2. What are the important topics of Class 11 Maths Chapter 10 Straight Lines?

Important topics of Class 11 Maths Chapter 10 Straight Lines:

Intercept form

Different forms of Ax + By + C = 0

3. Where can I find important questions for Class 11 Maths Chapter 10  Straight Lines?

Vedantu offers a well-prepared set of Important Questions for Class 11 Maths Chapter 10  Straight Lines as well as other chapters. Our teacher selects questions based on the exam pattern and questions from the previous papers. These solutions are also solved by subject matter experts to provide a thorough understanding of the chapter. These are definitely helpful in exam preparation and revisions during exams.

4. What are some of the important properties of straight lines?

Some of the important properties of straight lines are:

Two lines are perpendicular if the angle between them is 90°.

Two lines are parallel if the angle between them is 0°.

The product of the slopes of two perpendicular lines is equal to -1.

What are some of the applications of straight lines?

Straight lines have many applications in the real world, such as:

Building roads and bridges

Designing machines

Creating maps

Constructing buildings

5. How can I prepare for the Class 11 Maths exam on Straight Lines?

Here are some tips on how to prepare for the Class 11 Maths exam on Straight Lines:

Make sure you understand the basic concepts of straight lines.

Practice solving problems on straight lines.

Refer to the NCERT textbook , other standard textbooks and Vedantu’s notes

Take online tests and mock exams provided by vedantu.

6. How can we find the angle between two straight lines?

The angle between two straight lines can be defined as the angle between their direction vectors. The direction vector of a line is a vector that points in the direction of the line. The angle between two lines can be found using the dot product of their direction vectors.

CBSE Class 11 Maths Important Questions

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Course: Class 11   >   Unit 9

• Unit test Straight lines

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Case Study Questions Class 11 Physics Motion in a Straight Line

Case study questions class 11 physics chapter 3 motion in a straight line.

CBSE Class 11 Case Study Questions Physics Motion in a Straight Line. Important Case Study Questions for Class 11 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Motion in a Straight Line .

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

CBSE Case Study Questions Class 11 Physics Motion in a Straight Line

Case study – 1.

If an object moving along the straight line covers equal distances in equal intervals of time, it is said to be in uniform motion along a straight line. Distance and displacement are two quantities that seem to mean the same but are different with different meanings and definitions. Distance is the measure of actual path length travelled by object. It is scalar quantity having SI unit of metre while displacement refers to the shortest distance between initial and final position of object. It is vector quantity. The magnitude of the displacement for a course of motion may be zero but the corresponding path length is not zero. using this data answer following questions.

1) Can path length be zero for motion of body from one point to other point?

2) For any given motion from point A to B, displacement =10m and distance = 5m. Is it possible?

3) For rectilinear motion displacement can be

a) Positive only

b) Negative only

c) Can be zero

d) All of the above

4) Define distance and displacement of particle.

5) Write difference between distance and displacement.

Answer key – 1

4) Distance and displacement are two quantities that seem to mean the same but are different with different meanings and definitions. Distance is the measure of “how much distance an object has covered during its motion” while displacement refers to the measure of “how far the abject actually from initial place.”

5) difference between distance and displacement is given by

Case Study – 2 

When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. For convenience, a set of three such equations are given below:

v = u + at           s = ut + ½ at 2            2a s = v 2 – u 2

Where u is the initial velocity of the object which moves with uniform acceleration a for Time t, v is the final velocity and s is the distance travelled by the object in time t.

1) equation of motions are applicable to motion with

a) uniform acceleration

b) non uniform acceleration

c) constant velocity

d) none of these

2) There are 4 equation of motion. True or false?

3) The brakes applied to a car produce an acceleration of 10 m/s 2 in the opposite direction to the motion. If the car takes 1 s to stop after the application of brakes, calculate the distance traveled during this time by car.

4) An object is dropped from a tower falls with a constant acceleration of 10 m/s2. Find its speed 10 s after it was dropped.

5) A bullet hits a Sand box with a velocity of 10 m/s and penetrates it up to a distance of 5 cm. Find the deceleration of the bullet in the sand box

Answer key – 2

3) Here in this problem,

a = -10 m/s 2 (as acceleration is retarding)

To find: distance travelled

Solution: using kinematic equation

0=  u + -10*1

Therefore distance is given by

s = ut + ½ at 2

s = 10*1-(1/2)*10*1 2

4) here in this problem,

a= 10 m/s 2 (as acceleration is in the direction of gravity)

To find: final velocity after 10 second

v = 0+ 10*10

5) Here in this problem,

v= 0(as bullet is going to stop)

To find: deceleration of the bullet

2a s = v 2 – u 2

2*a*5= 0 2 -10 2

a = -10m/s 2 . Negative sign indicates that it is deceleration.

Case Study – 3

The average velocity tells us how fast an object has been moving over a given time interval but does not tell us how fast it moves at different instants of time during that interval. For this, we define instantaneous velocity or simply velocity v at an instant t. The velocity at an instant is defined as the limit of the average velocity as the time interval Dt becomes infinitesimally small. In other words

The variation in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph, time is represented along the x- axis and the velocity is represented along the y -axis. We know that the product of velocity and time give displacement of an object moving with uniform velocity. The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement and slope of velocity time graph represents acceleration of object.

The variation in distance with time for an object moving in a straight line can be represented by a position-time graph. In this graph, time is represented along the x- axis and the displacement is represented along the y -axis. Answer the following questions based on paragraph given.

1) The area under velocity time graph gives

a) Displacement over given time interval

b) Acceleration

c) Velocity

d) None of these

2) Slope of velocity time graph gives

a) Acceleration

b) Velocity

c) Distance

d) Displacement.

3) Write note on velocity time graph.

4) Write a note on position time graph

5) What is instantaneous velocity ?

Answer key-3

3) The change in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph, time is represented along the x- axis and the velocity is represented along the y -axis. slope of velocity time graph represents acceleration of object The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement and

4) The change in distance with time for an object moving in a straight line can be given by a position-time graph. In this graph, time is represented along the x- axis and the displacement is represented along the y -axis.

5) The velocity at an instant is defined as the limit of the average velocity as the time interval t becomes infinitesimally small. In other words

Case Study – 4

Introduce the concept of relative velocity.

Relative velocity is velocity of any object with respect to other object which may be stationary or moving. Consider two objects A and B moving uniformly with average velocities v A and v B in one dimension, say along x-axis. (Unless otherwise specified, the velocities mentioned in this chapter are measured with reference to the ground). If x A (0) and x B (0) are positions of objects A and B, respectively at time t = 0, their positions x A (t) and x B (t) at time t are given by

x A (t) = x A (0) + v A t

x B (t) = x B (0) + v B t

Then, the displacement from object A to object B is given by

x BA (t) = x B (t) – x A (t)

= [x B (0) – x A (0) ] + (v B – v A ) t.

It tells us that as seen from object A, object B has a velocity v B – v A because the displacement from A to B changes steadily by the amount v B – v A in each unit of time. We say that the velocity of object B relative to object A is v B – v A

V BA = v B – v A

Similarly, velocity of object A relative to object B is:

V AB = v A – v B

This shows V BA = – V AB .

1) Velocity of object A relative to object B is

a) V AB = v A – v B

b) V BA = v B – v A

c) None of these

2) Velocity of object B relative to object A is

A) v B – v A

b) v A – v B

3) What is relative velocity?

4) What is relative displacement?

5) Show that V BA = – V AB

Answer key-4

3) Relative velocity is velocity of any object with respect to other object which may be stationary or moving.

4) If x A (0) and x B (0) are positions of objects A and B, respectively at time t = 0, their positions x A (t) and x B (t) at time t are given by

x BA (t) = x B (t) – x A (t).

5) By definition of relative velocity We say that the velocity of object B relative to object A is v B – v A

Case Study – 5

When an object is in motion, its position changes with time. But how fast is the position changing with time and in what direction? To describe this, we define the quantity average velocity. Average velocity is defined as the change in position or displacement ( △ x) divided by the time intervals ( △ t), in which the displacement occurs:

Where x2 and x1 are the positions of the object at time t2and t1, respectively. The SI unit for velocity is m/s or m s –1 , although km h –1 is used in many everyday applications. Like displacement, average velocity is also a vector quantity. Average speed is defined as the total path length travelled divided by the total time interval during which the motion has taken place:

Average speed = Total path length/ Total time interval.

Average speed has obviously the same unit (m s –1 ) as that of velocity. But it does not tell us in what direction an object is moving. Thus, it is always positive (in contrast to the average velocity which can be positive or negative). If the motion of an object is along a straight line and in the same direction, the magnitude of displacement is equal to the total path length.

The velocity at an instant is defined as the limit of the average velocity as the time interval Dt becomes infinitesimally small. In other words

Note that for uniform motion, velocity is the same as the average velocity at all instants. Instantaneous acceleration is defined in the same way as the instantaneous velocity

1) For uniform motion instantaneous velocity is same as

a) Average velocity

b) Average acceleration

c) Instantaneous speed

2 If velocity is constant then

a) Acceleration is zero

Acceleration is positive

c) Acceleration is negative

3) Define average speed

4) Define instantaneous acceleration

5) Define average velocity

Answer key-5

3) Average speed is defined as the total path length travelled divided by the total time.

Average speed= Total path length / Total time interval.

Average speed has SI unit of m/s. it is scalar quantity it has only magnitude and doesn’t have any direction. it is always positive.

4) Instantaneous acceleration is defined rate of change of velocity with time when time tends to zero

5) Average velocity is defined as the change in position or displacement (Dx) divided by the time intervals (Dt), in which the displacement occurs :

Where x2 and x1 are the positions of the object at time t2 and t1 , respectively. The SI unit for velocity is m/s or m s –1 .

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• Mathematics /

Class 11 Straight Lines

• Updated on  
• Dec 10, 2020

The chapter 10 Straight Lines will be your dose for mathematical geometry as per class 11 maths syllabus . As geometry has been an integral part of the curriculum since junior classes, the same is even followed in class 11th and 12th. You must have been through the graphical representation of equations via chapters like linear equation in one variable . Thus, class 11 straight lines chapter will equip you with advanced knowledge of this concept. So let’s go through our notes and understand the chapter. 

This Blog Includes:

What is a straight line, slope of a line, horizontal and vertical lines, slope-intercept form, intercept form, distance between two parallel lines, straight lines: practice questions.

Any line that can be traced by a point travelling in a direction with zero curvature is known as a straight line. A straight line extends in two directions forever while having only one dimension and length. It is an important concept of geometry that is useful in various ways.

To represent a line algebraically, the slope is an essential part because, in a coordinate plane, it forms two angles with the x-axis in supplementary form. As per class 11 straight lines chapter, the slope of a line can be determined with the value of tan θ . Wherein, within the positive direction of the line in an anti-clock direction, an angle is made by θ. 

If the slope of the line passes through two points, i.e., P (a1, b1) and Q (a1, b2), then it becomes: m = tan θ = b2-b1/a2-a1.

Let us understand this through an example, let P (a1, b1) and Q (a1, b2) be any two points on a non-vertical line. The inclination of these lines will be θ , and it will be acute or obtuse. Under both the scenarios, the slope of the line will pass through point P and Q only.

Take This Maths Quiz If You Consider Yourself Genius!

Various Forms of the Equation of a Line

Class 11 maths NCERT states that every line in a plane has infinitely many points . By using the relationship between the line and points, several equations can be solved. Below mentioned are different conditions that explain the equations of a line. 

Derived form class 11 straight lines under this condition, if a horizontal line is distanced from a point to the x-axis, then every point will ordinate either positive or negative respectively, i.e., +a or -a. Similarly, it will be same for the vertical line also, but from a different axis than x-axis, i.e., y-axis.

Point-Slope Form Let P0 (x0, y0) be a fixed point on a non-vertical line with assumed slope “m”. If P (x, y) is an arbitrary point on the non-vertical line then, the point (x, y) will lie with slope “m” on the same line through the fixed point (x0, y0) with equation y – y0 = m (x – x0).

Two-Point Form If an imaginary line “L” passes through any two given points, i.e. P1 (x1, y1) and P2 (x2, y2) then, with general point P (x, y), all the three points, i.e. P1, P2, and P3 will be collinear, and the equation will be y−y1 = y2−y1 / x2−x1 * (x−x1).

Slope-Intercept Form Let us assume that a line cuts the slope from the y-axis at a distance “c”, then the distance “c” will be called a y-intercept of the line. When the line meets the y-axis, the coordinates will be (0, c) and the equation will be y = mx + c wherein the “c” can be positive or negative as mentioned in class 11 straight lines. If the line makes an x-intercept of the line instead of the y-intercept of the line, then the equation will be y = m (x-d).

Intercept-Form If a line makes “x-intercept a” and “y-intercept b” on any axis and the line meets x-axis at point (a, 0) and y-axis at point (0, b), then the equation according to two forms will be x/a + y/b = 1.Go through the class 11 maths NCERT solutions and get a good grip over the topic.

General Equation of a Line

A general equation of first degree in two variables will always be Ax + By + C = 0 . Wherein A, B, and C will be real constants with A and B a non-zero value. Graph of such an equation will always be a straight line, and therefore the equation Ax + By + C = 0 is called a general linear equation or general equation of a line. According to class 11 straight lines chapter, the general equation can be reduced into several other forms of equations by using the slope-intercept form, intercept form, and normal form.

If B is not equals to 0, then the equation Ax + By + C = 0 can be rewritten as y = mx + c. If B equals to zero, then x = -c/a.

If C is not equals to 0, then the equation Ax + By + C = 0 can be rewritten as x/a + y/b = 1. If C equals to 0, then the equation Ax + By + C = 0 can be rewritten as Ax + By = 0.

The slopes of two parallel lines will always be equal is one of the most essential concepts that are mentioned in class 11 straight lines chapter. If we take two parallel lines in the form of y = mx + c1 and y = mc + c2 , then the distance between the two lines will be equal to the length of the perpendicular. The equation for the two lines can be written as Ax + By + C1 = 0 and Ax + By + C2 = 0 .

Let us have a look at solved example-

Q: If points (x, 1), (2, 1), and (4, 5) are collinear, then find the value of x . Solution: The slope of AB = Slope of BC (1+1) / (2-x) = (5-1) / (4-2) 2 / (2-x) = 2 2 = 2 * (2-x) 2 = 4 – 2x 2x = 2 x = 2 / 2 x = 1

Here are some practices questions based on the similar concepts as mentioned in class 11 straight lines chapter.

• In the given equation 2x + y – 3 =0, 5x + ky – 3 = 0 and 3x – y – 2 = 0, if all the three lines are concurrent, then find the value of k.
• If a line is passing through the point (2, 3) with an angle of 60° from the x-axis, then find the equation of the line.
• What will be the distance of a point from a line?
• Find the distance between parallel lines 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.
• How coordinated geometry and analytical geometry are related to each other?
• Find the measure of the angle in between the lines x+y+7=0 and x-y+1=0.
• Find the equation of the line, which makes intercepts -3 and 2 on the x and y-axis respectively.
• Find the points on the x-axis whose distance from the line equation (x/3) + (y/4) = 1 is given as 4units. 
• At what point is the origin shifted, if the coordinates of the point (,5)become (3,7)?
•  Find the intercepts cut off by the line 2a-b+16=0
•  Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
•  Determine x so that the inclination of the line containing the points (x,-3) and (2,5) is 135.
•  Find the equation of the line which passes through the point (3,4) and the sum of whose intercepts on the axes is 14. 
•  Using slopes, find the value of x for which the points (x,-1) (2,1) and (4,5) are collinear.
•  Find the value of K, given that the distance of the point (4,1) from the line 3x-4y+k=0 IS 4 units.

Thus, we hope that through this blog that aims to explain the core fundamentals of class 11 straight lines with informative examples, you have understood the chapter in an easy to understand way. If you are clueless about how to proceed in the right direction leading towards your career goals, reach out to our career experts at Leverage Edu and they will guide you the best. Book an e-meeting now!

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Class 11 Maths Chapter 10 Straight Lines MCQs

Students can access Class 11 Maths Chapter 10 Straight Lines MCQs online, which will help them to score good marks in the examination. The multiple-choice questions are based on the CBSE syllabus (2022-2023) and NCERT guidelines. These multiple-choice questions have detailed answers and explanations. To obtain all chapter-by-chapter MCQs for Class 11 Maths, click here .

MCQs on Class 11 Maths Chapter 10 Straight Lines

Solve the MCQs for Class 11 Maths Chapter 10 Straight Lines. There are four options for each MCQ, but only one is accurate.

Download PDF – Chapter 10 Straight lines MCQs

Students should select the appropriate choice and compare their answers to those on our website. Also, look over the important questions for class 11 Maths .

1) Two lines are said to be perpendicular if the product of their slope is equal to:

Answer: (a) -1

Explanation:

When two lines are perpendicular, then the product of their slope is equal to -1. If two lines are perpendicular with slope m 1 and m 2 , then m 1 .m 2 = -1.

2) What is the distance of (5, 12) from the origin?

Answer: (d) 13 units

Explanation: Let the points be A(0, 0) and B(5, 12).

A (0, 0) = (x 1 , y 1 )

B(5, 12) = (x 2 , y 2 )

AB = √(25+144)

AB = √(169)

Hence, the distance of (5, 12) from the origin is 13 units.

3) Two lines are said to be parallel if the difference of their slope is

• None of these

Answer: (b) 0

We know that two lines are said to be parallel if their slope is equal. If m 1 and m 2 are the slopes of two parallel lines, then it is represented as m 1 = m 2 .

Hence, the difference of their slope should be m 1 -m 2 = 0.

So, option (b) 0 is the correct answer.

4) The equation of a straight line that passes through the point (3, 4) and perpendicular to the line 3x+2y+5=0 is

• 2x-3y+6 = 0
• 2x+3y+6 = 0
• 2x-3y-6 = 0
• 2x+3y-6 = 0

Answer: (a) 2x-3y+6 = 0

The equation of a straight line perpendicular to 3x+2y+5 = 0 is 2x-3y+λ =0 …(1)

This passes through the point (3, 4).

Now, substitute in equation (1), we get

2(2) – 3(4) +λ = 0

Substituting λ=6 in (1), we get 2x-3y+6 = 0, which is the required equation.

Hence, option (a) 2x-3y+6 =0 is the correct answer.

5) The slope of a line ax+by+c =0 is

Answer: (b) -a/b

We know that the general equation of a line is ax+by+c = 0.

Rearranging the equation, we get

⟹ by = -ax -c

⟹y = (-a/b)x -(c/b) …(1)

This is of the form, y= mx+c …(2)

By comparing (1) and (2), we get

Slope, m = -a/b.

6) The equation of a line that passes through the points (1, 5) and (2, 3) is:

• 2x + y – 7 = 0
• 2x – y – 7 = 0
• x + 2y – 7 = 0
• 2x + y + 7 = 0

Answer: (a) 2x + y – 7 = 0

We know that the equation of a line passes through two points (x 1 , y 1 ) and (x 2 y 2 ) is

(y-y 1 )/(x-x 1 ) = (y 2 -y 1 )/(x 2 -x 1 )

(x 1 , y 1 ) = (1, 5)

(x 2 , y 2 ) = (2, 3)

Now, substitute the values in the formula, we get

(y-5)/(x-1) = (3-5)/(2-1)

(y-5)/(x-1) = (-2)/(1)

y-5 = -2(x-1)

y-5 = -2x+2

2x+y-5-2 =0

Therefore, the equation of a line that passes through the points (1, 5) and (2, 3) is 2x+y-7=0.

7) The distance between the lines 3x+4y=9 and 6x+8y=15 is

Answer: (c) 3/10

Given equations:

3x + 4y = 9 …(i)

6x + 8y =15 …(ii)

⟹3x + 4y = 15/2

The slope of line (i) is -3/4.

The slope of line (ii) is -3/4.

Since slopes are equal, the lines are parallel.

Hence, the distance between two parallel lines = |(c 1 – c 2 )|/√(a 2 + b 2 )

= |(9 – (15/2))|/√(3 2 + 4 2 )

Therefore, the distance between the lines 3x+4y=9 and 6x+8y=15 is 3/10.

8) The locus of a point, whose abscissa and ordinate are always equal is

• None of the above

Answer: (a) x-y =0

Let the abscissa and ordinate of a point “P” be(x, y)

Given condition: Abscissa = Ordinate

Hence, the locus of a point is x-y = 0.

Therefore, option (a) x-y =0 is the correct answer.

9) If A(6, 4) and B(2, 12) are the two points, then the slope of a line perpendicular to line AB is

Answer: (c) ½

Given points: A(6, 4) = (x 1 , y 1 )

B(2, 12) = (x 2 , y 2 )

We know that the slope of a line passing through two points (x 1 , y 1 ) and (x 2 , y 2 ) is (y 2 -y 1 )/(x 2 -x 1 ).

m = (12-4)/(2-6) = 8/-4 = -2.

We know that the slope of two perpendicular lines m 1 .m 2 = -1.

So, m 2 = -1/m 1

Hence, the slope of a line perpendicular to line AB is -1/m = -1/-2 = ½.

10) What can be said regarding a line if its slope is negative?

• θ is an acute angle
• θ is an obtuse angle
• Either the line is x-axis or it is parallel to the x-axis.

Answer: (b) θ is an obtuse angle

The line with a negative slope makes an obtuse angle with a positive x-axis when measured in the anti-clockwise direction.

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Class 11 Physics Case Study Questions Chapter 3 Motion in a Straight Line

• Post author: studyrate
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In Class 11 Final Exams there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Downloads of CBSE Class 11 Physics Chapter 3 Case Study and Passage-Based Questions with Answers were Prepared Based on the Latest Exam Pattern. Students can solve Class 11 Physics Case Study Questions Motion in a Straight Line  to know their preparation level.

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In CBSE Class 11 Physics Paper, There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Motion in a Straight Line Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 11 Physics  Chapter 3 Motion in a Straight Line

Case Study/Passage-Based Questions

Case Study 1:

When an object is in motion, its position changes with time. But how fast is the position changing with time and in what direction? To describe this, we define the quantity average velocity. Average velocity is defined as the change in position or displacement ( △ x) divided by the time intervals ( △ t), in which the displacement occurs:

Where x2 and x1 are the positions of the object at times t2 and t1, respectively. The SI unit for velocity is m/s or m s–1, although km h–1 is used in many everyday applications. Like displacement, average velocity is also a vector quantity. Average speed is defined as the total path length traveled divided by the total time interval during which the motion has taken place:

Average speed = Total path length/ Total time interval.

Average speed has obviously the same unit (m s –1 ) as that of velocity. But it does not tell us in what direction an object is moving. Thus, it is always positive (in contrast to the average velocity which can be positive or negative). If the motion of an object is along a straight line and in the same direction, the magnitude of displacement is equal to the total path length.

The velocity at an instant is defined as the limit of the average velocity as the time interval Dt becomes infinitesimally small. In other words

Note that for uniform motion, velocity is the same as the average velocity at all instants. Instantaneous acceleration is defined in the same way as the instantaneous velocity

1) For uniform motion instantaneous velocity is the same as

• a) Average velocity
• b) Average acceleration
• c) Instantaneous speed
• d) None of these

Answer: a) Average velocity

2 If velocity is constant then

• a) Acceleration is zero
• b) Acceleration is positive
• c) Acceleration is negative

Answer: a) Acceleration is zero

What is average velocity defined as? a) Total path length / Total time interval b) Displacement / Time interval c) Speed / Direction d) Time interval / Displacement

Answer: b) Displacement / Time interval

Which unit is used for velocity in everyday applications? a) m/s b) m s–1 c) km/h d) Both a and b

Answer:c) km/h

Which of the following is true about average speed? a) It is a vector quantity b) It can be negative c) It is always positive d) It describes the direction of motion

Answer: c) It is always positive

Case Study 2:

When an object is in motion, its position changes with time. So, the quantity that describes how fast is the position changes w.r.t. time and in what direction is given by average velocity. It is defined as the change in position or displacement (Δx ) divided by the time interval (Δt ) in which that displacement occurs. However, the quantity used to describe the rate of motion over the actual path is average speed. It is defined as the total distance traveled by the object divided by the total time taken.

(i) A 250 m long train is moving with a uniform velocity of 45 km/h. The time taken by the train to cross a bridge of length 750 m is (a) 56 s (b) 68 s (c) 80 s (d) 92 s

Answer: (c) 80 s

(ii) A truck requires 3 hr to complete a journey of 150 km. What is the average speed? (a) 50 km/h (b) 25 km/h (c) 15 km/h (d) 10 km/h

Answer: (a) 50 km/h

(iii) Average speed of a car between points A and B is 20 m/s, between B and C is 15 m/s and between C and D is 10 m/s. What is the average speed between A and D, if the time taken in the mentioned sections is 20s, 10s and 5s, respectively? (a) 17.14 m/s (b) 15 m/s (c) 10 m/s (d) 45 m/s

Answer: (a) 17.14 m/s

(iv) A cyclist is moving on a circular track of a radius of 40 m and completes half a revolution in 40 s. Its average velocity (in m/s) is (a) zero (b) 2 (c) 4π (d) 8π

Answer: (b) 2

Hope the information shed above regarding Case Study and Passage Based Questions for Class 11 Physics Chapter 3 Motion in a Straight Line with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 11 Physics Motion in a Straight Line Case Study and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

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