Straight Lines — Class 11 MCQs with Answers

Straight Lines — important questions & MCQs with answers (Class 11 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Slope of line through (1,2) and (3,8):

   A3B2C4D6
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   Answer: 3

   Explanation: Slope = (y₂−y₁)/(x₂−x₁) = (8−2)/(3−1) = 3. Flipping which point is first, or dividing Δx by Δy instead of Δy by Δx, is the usual slip that lands on a wrong value like 2 or 4.

2. Q2 of 6Easy

   Slope-intercept form of a line:

   Ay = mx + cBax + by + c = 0Cy = xDx + y = 0
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   Answer: y = mx + c

   Explanation: y = mx + c is the slope-intercept form, reading off slope m and y-intercept c directly. The general form ax + by + c = 0 hides both behind extra algebra — rewrite it as y = mx + c to compare.

3. Q3 of 6Medium

   Slope of line through (1, 2) and (3, 4):

   A−1B2C1D0
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   Answer: 1

   Explanation: Slope = (4 − 2)/(3 − 1) = 2/2 = 1. Subtracting in the wrong order for just one coordinate — say (2−4)/(3−1) — flips the sign and gives −1 instead.

4. Q4 of 6Medium

   Assertion (A): The distance of the point (3, −5) from the line 3x − 4y − 26 = 0 is 7 units. Reason (R): The distance of the point (x₁, y₁) from the line Ax + By + C = 0 is |Ax₁ + By₁ + C| / √(A² + B²).

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: A is false but R is true

   Explanation: Using R: |3(3) − 4(−5) − 26| / √(9 + 16) = |9 + 20 − 26| / 5 = 3/5 units, not 7. So A is false and R is true.

5. Q5 of 6Hard

   Acute angle between lines 3x − 4y + 7 = 0 and 4x + 3y − 5 = 0:

   A60°B90°C45°D30°
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   Answer: 90°

   Explanation: Slopes are 3/4 and −4/3; their product is (3/4)(−4/3) = −1, which is exactly the perpendicular condition, so the angle is 90°. Missing that this product equals −1 leads to computing a non-right angle instead.

6. Q6 of 6Hard

   Assertion (A): The image of the point (1, 2) in the line x − 3y + 4 = 0 is (6/5, 7/5). Reason (R): The image of a point P in a line is the point Q such that the segment PQ is parallel to that line.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: A is true but R is false

   Explanation: For Q(6/5, 7/5): slope of PQ = (7/5 − 2)/(6/5 − 1) = −3, which is perpendicular to the mirror of slope 1/3, and the mid-point (11/10, 17/10) satisfies 11/10 − 51/10 + 4 = 0. So A is true. R is false: the mirror line is the perpendicular bisector of PQ, so PQ is perpendicular to it.

Straight Lines notes: key concepts (Class 11 Mathematics)

Coordinate geometry turns lines into equations. This chapter defines the slope of a line as the tangent of its inclination, derives the slope-intercept, point-slope, two-point, intercept and normal forms, reduces the general equation Ax + By + C = 0, and finds the angle between two lines and the distance of a point from a line.

Slope of a line

m = (y₂ − y₁)/(x₂ − x₁) = tan θ, where θ is the angle the line makes with the positive x-axis; a vertical line has undefined slope.

Inclination

The angle θ, with 0° ≤ θ < 180°, measured anticlockwise from the positive x-axis to the line; its tangent gives the slope.

Parallel and perpendicular

Parallel lines share equal slopes m₁ = m₂; perpendicular lines satisfy m₁m₂ = −1, so each slope is the negative reciprocal of the other.

Slope-intercept form

y = mx + c, where m is the slope and c the y-intercept; the fastest form when the slope and the y-axis crossing are known.

Point-slope form

y − y₁ = m(x − x₁) writes the line through a known point (x₁, y₁) with slope m; the natural choice given a point and a slope.

Two-point form

y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁) builds the line through two given points by first forming the slope between them.

Intercept form

x/a + y/b = 1, where a and b are the x- and y-intercepts; handy when a line's two axis crossings are given directly.

Normal form

x cos ω + y sin ω = p, where p is the perpendicular distance from the origin and ω the angle that perpendicular makes with the x-axis.

General equation

Ax + By + C = 0 represents every straight line; its slope is −A/B, its x-intercept −C/A and its y-intercept −C/B.

Angle between two lines

tan θ = |(m₁ − m₂)/(1 + m₁m₂)| gives the acute angle; lines are parallel if m₁ = m₂ and perpendicular if 1 + m₁m₂ = 0.

Distance of a point from a line

The perpendicular distance from a point (x₁, y₁) to the line Ax + By + C = 0 equals |Ax₁ + By₁ + C|/√(A² + B²).

Distance between parallel lines

For Ax + By + C₁ = 0 and Ax + By + C₂ = 0 the gap is |C₁ − C₂|/√(A² + B²), using matched coefficients A and B.

Image of a point in a line

The reflection of a point in a line sits on the perpendicular through it, equally far on the opposite side; the foot of that perpendicular is the midpoint of the point and its image.