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Chapter 13 · Class 11 Mathematics

Straight Lines — Important Questions

33 questions With answers CBSE format

SUMMARY: The chapter on Straight Lines in Class 11 Mathematics explores the concepts and equations related to lines in a two-dimensional plane.
KEY TOPICS: slope of a line, point-slope form, slope-intercept form, general form of a line, distance of a point from a line, angle between two lines, parallel and perpendicular lines, collinearity of points, intersection of lines, family of lines

Q1 1 Mark

The slope of the line passing through (1, 2) and (3, 6) is:

A1
B2
C4
D1/2
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Correct answer: Option 2 — 2
Q2 1 Mark

The equation of the line with slope 3 passing through (2, 1) is:

Ay = 3x + 1
By = 3x − 5
Cy − 1 = 3(x − 2)
Dy + 1 = 3(x + 2)
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Correct answer: Option 3 — y − 1 = 3(x − 2)
Q3 1 Mark

Two lines y = m₁ x + c₁ and y = m₂ x + c₂ are perpendicular iff:

Am₁ + m₂ = 0
Bm₁ m₂ = −1
Cm₁ = m₂
Dm₁ m₂ = 1
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Correct answer: Option 2 — m₁ m₂ = −1
Q4 1 Mark

The distance from origin to the line 3x + 4y − 5 = 0 is:

A1
B5
C5/√7
D1/5
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Correct answer: Option 1 — 1
Q5 1 Mark

The line passing through (0, 3) and (2, 0) has slope:

A3/2
B−3/2
C2/3
D−2/3
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Correct answer: Option 2 — −3/2
Q6 3 Marks

Find the slope of the line joining the points (1, 3) and (4, 9).

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Slope m = (9 − 3)/(4 − 1) = 6/3 = 2.
Q7 3 Marks

Write the equation of the line in slope-intercept form with slope −2 and y-intercept 5.

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Slope-intercept form: y = mx + c. Here m = −2 c = 5 so y = −2x + 5.
Q8 3 Marks

Find the equation of the line passing through the points (1, 2) and (3, 8).

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Slope = (8 − 2)/(3 − 1) = 6/2 = 3. Using point (1 2): y − 2 = 3(x − 1) ⇒ y = 3x − 1.
Q9 3 Marks

Find the perpendicular distance from the origin to the line 3x + 4y = 12.

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Distance = |3·0 + 4·0 − 12|/√(3² + 4²) = 12/5 = 2.4 units.
Q10 3 Marks

Convert the equation y = 3x + 4 into general form.

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y = 3x + 4 ⇒ 3x − y + 4 = 0. In general form Ax + By + C = 0: A = 3 B = −1 C = 4.
Q11 6 Marks

Find the equation of the line passing through the point of intersection of the lines x + y = 5 and 2x − y = 1, and parallel to the line 3x + 4y = 7.

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Solve x + y = 5 and 2x − y = 1: adding gives 3x = 6 ⇒ x = 2 and y = 3. So intersection point is (2, 3). Line 3x + 4y = 7 has slope −3/4. Parallel line through (2, 3): y − 3 = (−3/4)(x − 2) ⇒ 3x + 4y = 18.
Q12 6 Marks

Show that the points (1, 2), (3, 6) and (5, 10) are collinear.

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Slope of segment 1: (6 − 2)/(3 − 1) = 4/2 = 2. Slope of segment 2: (10 − 6)/(5 − 3) = 4/2 = 2. Slopes equal so the three points are collinear.
Q13 6 Marks

Find the equation of the line passing through (3, 4) and perpendicular to 2x − 3y + 5 = 0.

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Slope of 2x − 3y + 5 = 0 is 2/3. Perpendicular slope = −3/2. Required line: y − 4 = (−3/2)(x − 3) ⇒ 2y − 8 = −3x + 9 ⇒ 3x + 2y = 17.
Q14 6 Marks

Find the angle between the lines y = 2x + 3 and y = 3x − 1.

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Slopes m₁ = 2 m₂ = 3. tan θ = |(m₂ − m₁)/(1 + m₁ m₂)| = |1/(1 + 6)| = 1/7. So θ = tan⁻¹(1/7) ≈ 8.13°.
Q15 6 Marks

Find the area of the triangle formed by the lines y = 0, x = 4 and y = x + 2.

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Vertices: y = 0 and y = x + 2 meet at (−2, 0); x = 4 and y = 0 meet at (4, 0); x = 4 and y = x + 2 meet at (4, 6). Triangle with vertices (−2, 0), (4, 0), (4, 6) has base 6 (along x-axis from −2 to 4) and height 6. Area = (1/2)(6)(6) = 18 sq units.
Q16 6 Marks

Compare slope-intercept form and point-slope form of a straight line with the help of a table.

Q17 1 Mark

Assertion (A): Slope of a line through (x₁ y₁) and (x₂ y₂) is (y₂ − y₁)/(x₂ − x₁) for x₁ ≠ x₂.

Reason (R): Slope measures the rate of change of y with respect to x.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q18 1 Mark

Assertion (A): Two non-vertical lines are parallel iff their slopes are equal.

Reason (R): Parallel lines never meet so they share the same direction (slope).

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q19 1 Mark

Assertion (A): Two non-vertical lines are perpendicular iff the product of their slopes is −1.

Reason (R): The negative reciprocal relation reflects that perpendicular lines meet at 90°.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q20 1 Mark

Assertion (A): The equation of the x-axis is y = 0.

Reason (R): Every point on the x-axis has y-coordinate equal to 0.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q21 1 Mark

Assertion (A): A line with x-intercept a and y-intercept b satisfies x/a + y/b = 1.

Reason (R): The line passes through (a 0) and (0 b) and the equation can be derived by the two-point form.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q22 1 Mark

Statement 1: The equation y = mx + c has slope m.

Statement 2: The equation y = mx + c has y-intercept c.

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Correct answer: Option 1 — Both statements are true.
Q23 1 Mark

Statement 1: A horizontal line has slope 0.

Statement 2: A vertical line has undefined slope.

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Correct answer: Option 1 — Both statements are true.
Q24 1 Mark

Statement 1: Two lines with slopes 2 and −1/2 are perpendicular.

Statement 2: Their slopes have product −1.

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Correct answer: Option 1 — Both statements are true.
Q25 1 Mark

Statement 1: The point-slope form of a line is y − y₁ = m(x − x₁).

Statement 2: The two-point form generalises the point-slope form using two known points.

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Correct answer: Option 1 — Both statements are true.
Q26 1 Mark

Statement 1: A line with x-intercept 4 and y-intercept 3 has equation x/4 + y/3 = 1.

Statement 2: An equation in intercept form passes through (a 0) and (0 b).

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Correct answer: Option 1 — Both statements are true.
Q27 3 Marks
A straight road passes through the points P(2, 3) and Q(5, 7). A bridge perpendicular to the road must be built at the point R(4, 5).
  1. The slope of the road equals:
    A3/4
    B4/3
    C1
    D−1
  2. The slope of the bridge (perpendicular) equals:
    A4/3
    B−3/4
    C3/4
    D−4/3
  3. Write the equation of the bridge in slope-intercept and general forms.
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1. Option 2 — 4/3
2. Option 2 — −3/4
3. Slope of PQ = (7 − 3)/(5 − 2) = 4/3. Perpendicular slope = −1/(4/3) = −3/4. Equation of bridge through R(4, 5): y − 5 = −3/4 (x − 4) ⇒ 4y + 3x = 32.
Q28 3 Marks
A triangular plot has vertices A(0, 0), B(4, 0) and C(0, 3). The owner wants the area equations of the three sides and the slope of each side.
  1. The area of the triangle equals:
    A3
    B4
    C6
    D12
  2. The slope of side BC equals:
    A3/4
    B−3/4
    C4/3
    D−4/3
  3. Write the equation of side BC in intercept form.
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1. Option 3 — 6
2. Option 2 — −3/4
3. Area = (1/2)·base·height = (1/2)·4·3 = 6 sq units. Slope AB = 0 (horizontal). Slope AC = undefined (vertical). Slope BC = (3 − 0)/(0 − 4) = −3/4. Equation BC: x/4 + y/3 = 1.
Q29 3 Marks
A surveyor wants to measure the perpendicular distance from the point (3, −2) to the line 4x − 3y + 5 = 0.
  1. The distance from (3, −2) to the line equals:
    A5
    B5/√7
    C23/5
    D5/2
  2. The formula |Ax₀ + By₀ + C|/√(A² + B²) gives:
    APerpendicular distance from a point to a line
    BDistance between two parallel lines
    CDistance between two intersecting lines
    DLength of the line
  3. Compute the distance step by step.
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1. Option 3 — 23/5
2. Option 1 — Perpendicular distance from a point to a line
3. d = |4(3) − 3(−2) + 5|/√(4² + (−3)²) = |12 + 6 + 5|/√25 = 23/5 = 4.6 units.
Q30 3 Marks

Study the slopes of various lines:

LineSlope
y = 2x + 32
x = 4undefined
y = −50
y = (1/2)x − 11/2
3x + 4y = 12−3/4
  1. The slope of the line y = 2x + 3 equals:
    A2
    B1/2
    C−3/4
    DUndefined
  2. The slope of the line x = 4 is:
    ADefined non-zero
    BDefined and zero
    CUndefined
    DNegative
  3. Find the slope of the line 5x − 2y + 7 = 0.
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1. Option 1 — 2
2. Option 3 — Undefined
3. Vertical lines have undefined slope. Horizontal lines have slope 0. Lines in slope-intercept form y = mx + c have slope m. Lines in general form Ax + By + C = 0 have slope −A/B (when B ≠ 0).
Q31 3 Marks

Study the equations of standard lines:

FormEquation
Slope-intercepty = mx + c
Point-slopey − y₁ = m(x − x₁)
Two-point(y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁)
Interceptx/a + y/b = 1
GeneralAx + By + C = 0
  1. The form y − y₁ = m(x − x₁) is the:
    ASlope-intercept
    BPoint-slope
    CTwo-point
    DIntercept
  2. The form x/a + y/b = 1 is the:
    ATwo-point
    BSlope-intercept
    CIntercept
    DGeneral
  3. Convert y − 2 = 3(x + 1) to general form.
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1. Option 2 — Point-slope
2. Option 3 — Intercept
3. Each form is suited to different given information. Slope-intercept needs slope and y-intercept; point-slope needs slope and any point; two-point needs two points; intercept needs both intercepts; general form is the most flexible.
Q32 6 Marks

From the points in the table, find the slope of each line segment and identify which two are parallel and which are perpendicular.

LinePoint 1Point 2
L1(1, 2)(4, 8)
L2(0, 3)(3, 9)
L3(2, 5)(4, 4)
Q33 3 Marks

Study the line y = 2x + 3 with slope-intercept marked and answer:

Straight Lines figure
  1. The slope of the line equals:
    A1
    B2
    C3
    D−2
  2. The y-intercept is at the point:
    A(0, 0)
    B(0, 2)
    C(0, 3)
    D(3, 0)
  3. State the slope-intercept form and identify m and c.
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1. Option 2 — 2
2. Option 3 — (0, 3)
3. Slope-intercept form y = mx + c immediately gives slope m and y-intercept c. Here y = 2x + 3 has slope 2 and y-intercept 3. Slope = rise/run = 2/1 — for every unit increase in x, y increases by 2 units.

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