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CBSE · Class 11 · Mathematics

Straight Lines

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The slope of the line through (−5, 2) and (8, 8) is:
  1. 13/6
  2. −6/13
  3. 19/13
  4. 6/13
2.
The equation of the line through (7, 6) with slope −2 is:
  1. x − 2y + 20 = 0
  2. 2x + y + 20 = 0
  3. 2x − y − 20 = 0
  4. 2x + y − 20 = 0
3.
A line has slope −5/4. The slope of a line perpendicular to it is:
  1. 4/5
  2. −5/4
  3. 5/4
  4. −4/5
4.
The intercepts of the line 2x − 3y − 12 = 0 on the x- and y-axes are:
  1. −2 and 3
  2. −4 and 6
  3. −6 and 4
  4. 6 and −4
5.
Two non-vertical lines are parallel if and only if their slopes are:
  1. both 1
  2. negative reciprocals
  3. both zero
  4. equal
6.
The distance of the point (1, −8) from the line 20x − 21y − 362 = 0 is:
  1. 6
  2. 12
  3. 7
  4. 174
7.
A line makes an angle of 30° with the positive direction of the x-axis. Its slope is:
  1. 0
  2. −1/√3
  3. −1
  4. 1/√3
8.
Assertion (A): A line with slope 3/5 is perpendicular to a line with slope −5/3.
Reason (R): Two lines are perpendicular when the product of their slopes is 1.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Find the equation of the line through (5, 7) and (8, 8).
10.
Find the equation of the line that makes intercepts 1 and 3 on the x- and y-axes.
11.
Find the distance of the point (5, 7) from the line 15x − 8y − 21 = 0.
12.
Find the distance between the parallel lines 12x + 5y − 13 = 0 and 12x + 5y + 8 = 0.
13.
Use slopes to decide whether the points (7, 0), (11, 0) and (23, 0) are collinear.
14.
Find the equation of the line through (0, −8) that is perpendicular to 5x − 4y − 3 = 0.
15.
Write 8x − 8y − 7 = 0 in slope-intercept form, and find its slope and its intercepts on the axes.
16.
Find the tangent of the acute angle between two lines with slopes 1 and 3/2.
Section C — Word problems4 × 4 = 16 marks
17.
The vertices of a triangle are P(5, −7), Q(1, 2) and R(6, −5). Find the equations of (a) side QR, (b) the median from P, (c) the altitude from P.
18.
Find the foot of the perpendicular from (5, 7) to the line 4x − 5y − 7 = 0, and the length of the perpendicular.
19.
Find the point of intersection of 4x − 3y + 24 = 0 and 2x − y + 12 = 0. Then find the line through this point with slope −4, and the line through it and the origin.
20.
Find the equation of the perpendicular bisector of the segment joining A(−8, 6) and B(−6, −2). Check that the midpoint lies on it.
AmbifulWorksheets
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CBSE · Class 11 · Mathematics

Answer key

Straight Lines · AW-11M-SLN-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 6/13(8 − 2) ÷ (8 − (−5))
2(d) 2x + y − 20 = 0y − 6 = −2(x − 7)
3(a) 4/5m1m2 = −1
4(d) 6 and −4put y = 0, then x = 0
5(d) equal—
6(a) 6|−20 × 1 + 21 × (−8) + 362| ÷ √(202 + 212) = 174 ÷ 29
7(d) 1/√3tan 30°
8(c) A is true, but R is false.A is true: 3/5 × −5/3 = −1. R is false.
9x − 3y + 16 = 0slope 1/3; y − 7 = m(x − 5)
103x + y − 3 = 0x/1 + y/3 = 1
112/17|−2| ÷ 17
1221/13|C1 − C2| ÷ √(A2 + B2) = 21 ÷ 13
13Slope AB = 0, slope BC = 0: equal, so they are collinear—
144x + 5y + 40 = 0slope −4/5 (negative reciprocal of 5/4)
15y = x − 7/8; slope 1; x-intercept 7/8, y-intercept −7/8y = (−a/b)x − c/b
16tan θ = 1/5tan θ = |(m2 − m1) ÷ (1 + m1m2)|
17(a) 7x + 5y − 17 = 0; (b) 11x + 3y − 34 = 0; (c) 5x − 7y − 74 = 0midpoint of QR (7/2, −3/2); altitude ⟂ QR
18Foot (293/41, 177/41); length 22√41/41foot = (x1 + ak, y1 + bk) with k = −(ax1 + by1 + c) ÷ (a2 + b2) = 22/41
19(−6, 0); 4x + y + 24 = 0; y = 0solve the two equations together
20x − 4y + 15 = 0; midpoint (−7, 2) satisfies itpoints equidistant from A and B: (x − (−8))2 + (y − 6)2 = (x − (−6))2 + (y − (−2))2