Maths · Coordinate geometry · chapter 9 of 14

Class 11 Maths Chapter 9: Straight Lines worksheet with answer key

A 20-mark worksheet on Straight Lines for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Download the worksheet (PDF)Answer key (PDF)
Question 1
The slope of a line whose inclination with the positive direction of the x-axis is 150° is
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Answer: (c) −1/√3

m = tan 150° = tan (180° − 30°) = −tan 30° = −1/√3.

Question 2
Which of the following lines has a slope that is not defined?
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Answer: (a) x = −3

x = −3 is a vertical line: its inclination is 90°, and tan 90° is not defined. y = 4 has slope 0; y = x has slope 1; x + y = 0 has slope −1.

Question 3
The equation of the line through (−3, 1) whose inclination is 45° is
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Answer: (c) x − y + 4 = 0

m = tan 45° = 1. Point-slope form: y − 1 = 1(x + 3), i.e. x − y + 4 = 0. x + y + 2 = 0 passes through (−3, 1) but has inclination 135°.

Question 4
The line through (2, −1) and (5, a) is parallel to the line 2x − 3y + 7 = 0. The value of a is
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Answer: (b) 1

2x − 3y + 7 = 0 has slope −A/B = 2/3. Parallel lines have equal slopes: (a + 1)/(5 − 2) = 2/3, so a + 1 = 2 and a = 1.

Question 5
The acute angle between the lines y = 3x + 1 and y = −2x + 4 is
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Answer: (a) 45°

m₁ = 3, m₂ = −2: tan θ = |(−2 − 3)/(1 + (3)(−2))| = |−5/−5| = 1, so θ = 45°. 135° is the obtuse angle.

Question 6 · 2 marks
Find the points on the x-axis that are equidistant from the lines 3x + 4y − 12 = 0 and 4x + 3y − 12 = 0.
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For (h, 0): |3h − 12|/5 = |4h − 12|/5. Either 3h − 12 = 4h − 12, giving h = 0, or 3h − 12 = −(4h − 12), giving 7h = 24, h = 24/7. The points are (0, 0) and (24/7, 0).

  • |3h − 12| = |4h − 12| (1)
  • (0, 0) and (24/7, 0) (1)
Question 7 · 2 marks
The lines L₁: y = 2x and L₂: y = −3x + 1 have inclinations α₁ and α₂. Find tan θ, where θ = α₂ − α₁, and say which of the two adjacent angles between the lines is acute.
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tan θ = (m₂ − m₁)/(1 + m₁m₂) = (−3 − 2)/(1 − 6) = 1. As this is positive, θ = 45° is acute and the adjacent angle φ = 180° − 45° = 135° is obtuse.

  • tan θ = 1 (1)
  • θ = 45° acute, φ = 135° obtuse (1)
Question 8 · 2 marks
Show that the lines x − 2y + 3 = 0, 2x + y − 4 = 0 and 3x − y − 1 = 0 are concurrent, and find the common point.
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From the first, x = 2y − 3; in the second, 4y − 6 + y − 4 = 0, so y = 2 and x = 1. In the third, 3(1) − 2 − 1 = 0, so (1, 2) lies on all three lines: they are concurrent at (1, 2).

  • intersection (1, 2) of two of the lines (1)
  • verifies the third line passes through it (1)
Question 9 · 3 marks
The line 3x − 4y + 12 = 0 meets the x-axis at P and the y-axis at Q. Find the area of the triangle OPQ, O being the origin, and the length of the perpendicular from O to the line. Check your answers against each other.
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At y = 0, x = −4: P(−4, 0). At x = 0, y = 3: Q(0, 3). Area = (1/2) × 4 × 3 = 6 square units. Perpendicular from O: |0 − 0 + 12|/√(9 + 16) = 12/5. Check: PQ = 5, and (1/2) × PQ × 12/5 = 6, the same area.

  • P(−4, 0), Q(0, 3) and area 6 (1)
  • perpendicular 12/5 (1)
  • check (1/2) × 5 × 12/5 = 6 (1)
Question 10 · 3 marks
Using the inclinations of its sides, show that the triangle with vertices A(0, 0), B(3, √3) and C(0, 2√3) is equilateral.
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AB has slope √3/3 = 1/√3, inclination 30°. AC lies on the y-axis, inclination 90°. BC has slope (2√3 − √3)/(0 − 3) = −1/√3, inclination 150°. Angle A = 90° − 30° = 60°; angle C, between inclinations 90° and 150°, is 60°. So angle B = 180° − 120° = 60° (check: tan B = |(−1/√3 − 1/√3)/(1 − 1/3)| = √3). All angles are 60°: equilateral.

  • inclinations 30°, 90°, 150° (1)
  • angles A and C are 60° (1)
  • angle B = 60°, so equilateral (1)
Question 11 · 3 marks
The vertices of a triangle are A(0, 0), B(6, 0) and C(0, 4). Find the equations of its three medians and show that they are concurrent.
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Mid-points: BC (3, 2), CA (0, 2), AB (3, 0). Median from A through (3, 2): 2x − 3y = 0. Median from B through (0, 2): x/6 + y/2 = 1, i.e. x + 3y = 6. Median from C through (3, 0): x/3 + y/4 = 1, i.e. 4x + 3y = 12. The first two meet where 3x = 6: (2, 4/3). In the third, 4(2) + 3(4/3) = 12, so all three pass through (2, 4/3).

  • the three mid-points and median equations (1)
  • intersection (2, 4/3) of two medians (1)
  • third median passes through it (1)
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