Straight Lines – Class 11 Maths Worksheet

The slope of the line passing through the points (3, –2) and (–1, 4) is:

MCQ

(A) 3/2

(B) –3/2

(D) –2/3

Answer: (B) –3/2
m = (y₂ – y₁)/(x₂ – x₁) = (4 – (–2))/(–1 – 3) = 6/(–4) = –3/2.

The slope of the line making an inclination of 60° with the positive x-axis is:

MCQ

(A) 1/2

(B) 1

(D) 1/√3

Answer: (C) √3
m = tan θ = tan 60° = √3.

Two lines are perpendicular to each other if and only if the product of their slopes is:

MCQ

(A) 0

(B) 1

(D) Undefined

Answer: (C) –1
Two non-vertical lines l₁ and l₂ are perpendicular if and only if m₁ × m₂ = –1. This is because if one line has slope m, its perpendicular has slope –1/m.

Line through (–2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (x, 24). Find x.

MCQ · Numerical

(A) –4

(B) 4

(D) –2

Answer: (B) 4
m₁ = (8–6)/(4–(–2)) = 2/6 = 1/3. m₂ = 12/(x–8). For perpendicularity: m₁ × m₂ = –1 → (1/3) × 12/(x–8) = –1 → x–8 = –4 → x = 4.

The equation of a line through (–2, 3) with slope –4 is:

MCQ · Numerical

(A) 4x + y + 5 = 0

(B) 4x – y + 5 = 0

(D) x – 4y – 5 = 0

Answer: (A) 4x + y + 5 = 0
Point-slope form: y – 3 = –4(x – (–2)) → y – 3 = –4x – 8 → 4x + y + 5 = 0.

The equation of the line passing through (1, –1) and (3, 5) is:

MCQ · Numerical

(A) 3x – y – 4 = 0

(B) –3x + y + 4 = 0

(D) x – 3y + 4 = 0

Answer: (B) –3x + y + 4 = 0
Slope = (5–(–1))/(3–1) = 6/2 = 3. Using point (1, –1): y – (–1) = 3(x – 1) → y + 1 = 3x – 3 → –3x + y + 4 = 0.

The equation of a line with y-intercept 2 and slope 1/2 is:

MCQ · Numerical

(A) y = 2x + 1/2

(B) y = x/2 + 2

(D) Both B and C

Answer: (D) Both B and C
Slope-intercept form: y = mx + c → y = (1/2)x + 2. Multiplying both sides by 2: 2y = x + 4. Both (B) and (C) are equivalent forms.

A line makes x-intercept –3 and y-intercept 2. Its equation in general form is:

MCQ · Numerical

(A) 3x – 2y + 6 = 0

(B) 2x – 3y + 6 = 0

(D) 3x + 2y + 6 = 0

Answer: (B) 2x – 3y + 6 = 0
Intercept form: x/a + y/b = 1 → x/(–3) + y/2 = 1 → multiply by –6 → 2x – 3y + 6 = 0.

The perpendicular distance of the point (3, –5) from the line 3x – 4y – 26 = 0 is:

MCQ · Numerical

(A) 1/5

(B) 3/5

(D) 3

Answer: (B) 3/5
d = |Ax₁ + By₁ + C| / √(A² + B²) = |3(3) + (–4)(–5) – 26| / √(9 + 16) = |9 + 20 – 26| / 5 = |3| / 5 = 3/5.

The distance between the parallel lines 3x – 4y + 7 = 0 and 3x – 4y + 5 = 0 is:

MCQ · Numerical

(A) 1/5

(B) 2/5

(D) 4/5

Answer: (B) 2/5
d = |C₁ – C₂| / √(A² + B²) = |7 – 5| / √(9 + 16) = 2/5.

If the angle between two lines is π/4 and slope of one line is 1/2, the slope of the other line is:

MCQ · Numerical

(A) 3 only

(B) –1/3 only

(D) 1/3 or –3

Answer: (C) 3 or –1/3
Using tan θ = |(m₂ – m₁)/(1 + m₁m₂)| = 1. With m₁ = 1/2: |(m – 1/2)/(1 + m/2)| = 1. This gives m – 1/2 = 1 + m/2 → m/2 = 3/2 → m = 3, or m – 1/2 = –1 – m/2 → 3m/2 = –1/2 → m = –1/3.

The equation of the line parallel to x-axis and passing through (–2, 3) is:

MCQ

(A) x = –2

(B) y = –2

(D) x = 3

Answer: (C) y = 3
A line parallel to the x-axis has equation y = constant. Since it passes through (–2, 3), the y-coordinate is 3, so the equation is y = 3.

The area of the triangle with vertices (4, 4), (3, –2) and (–3, 16) is:

MCQ · Numerical

(A) 24

(B) 27

(D) 54

Answer: (B) 27
Area = (1/2)|x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂)| = (1/2)|4(–2–16) + 3(16–4) + (–3)(4–(–2))| = (1/2)|–72 + 36 – 18| = (1/2)|–54| = 27.

If three lines 2x + y – 3 = 0, 5x + ky – 3 = 0, and 3x – y – 2 = 0 are concurrent, the value of k is:

MCQ · Numerical

(A) –2

(B) 2

(D) 3

Answer: (A) –2
Solving lines 1 and 3: 2x+y–3=0 and 3x–y–2=0 → 5x = 5 → x = 1, y = 1. Substituting (1,1) into line 2: 5(1) + k(1) – 3 = 0 → k = –2.

The distance between the points (6, –4) and (3, 0) is:

MCQ · Numerical

(A) 4 units

(B) 5 units

(D) 3 units

Answer: (B) 5 units
PQ = √((3–6)² + (0–(–4))²) = √(9 + 16) = √25 = 5 units.

Find the slope of a line passing through the origin and the midpoint of segment joining P(0, –4) and B(8, 0).

One Word

Answer: –1/2
Midpoint of PB = ((0+8)/2, (–4+0)/2) = (4, –2). Slope from origin (0,0) to (4,–2) = (–2–0)/(4–0) = –1/2.

Find the value of x if line through (–2, 6) & (4, 8) is perpendicular to line through (8, 12) & (x, 24).

One Word

Answer: 4
m₁ = 1/3, m₂ = 12/(x–8). Since m₁m₂ = –1: (1/3)(12/(x–8)) = –1 → 4/(x–8) = –1 → x – 8 = –4 → x = 4.

A line has y-intercept –3/2 and slope 1/2. What is the x-intercept of this line?

One Word

Answer: 3
y = (1/2)x – 3/2. Set y = 0: (1/2)x = 3/2 → x = 3.

Find the perpendicular distance of the point (3, –5) from the line 3x – 4y – 26 = 0.

One Word

Answer: 3/5
d = |3(3) + (–4)(–5) – 26| / √(9+16) = |9 + 20 – 26| / 5 = 3/5.

What is the distance between the parallel lines 3x – 4y + 7 = 0 and 3x – 4y + 5 = 0?

One Word

Answer: 2/5
d = |C₁ – C₂| / √(A² + B²) = |7 – 5| / √(9+16) = 2/5.

Find the slope of a line making an angle of 30° with the positive direction of the y-axis (measured anticlockwise).

One Word

Answer: –√3
Angle with y-axis = 30° → Angle with x-axis = 90° + 30° = 120°. Slope = tan 120° = –tan 60° = –√3.

What is the value of k if the point of intersection of lines 2x + y – 3 = 0 and 3x – y – 2 = 0 lies on line 5x + ky – 3 = 0?

One Word

Answer: –2
Intersection of first two lines: x = 1, y = 1. Substituting in third: 5 + k – 3 = 0 → k = –2.

A line makes intercepts –3 and 2 on the x and y axes. Find the value of 2x – 3y at the x-intercept point.

One Word

Answer: –6
Equation: x/(–3) + y/2 = 1 → 2x – 3y + 6 = 0. At x-intercept (–3, 0): 2(–3) – 3(0) = –6.

Find the slope of the line through (3, –2) and (3, 4). (Write "undefined" or the number)

One Word

Answer: Undefined
m = (4–(–2))/(3–3) = 6/0, which is undefined. The line is vertical (parallel to y-axis), with equation x = 3.

Find the distance between points P(4, 1) and Q(1, 4) (the intersection points used in finding the distance of line 4x – y = 0 from P along a 135° line).

One Word

Answer: 3√2
PQ = √((4–1)² + (1–4)²) = √(9 + 9) = √18 = 3√2 units.