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Straight Lines – Class 11 Mathematics notes, important questions and MCQs

NCERT Class 11 Mathematics, chapter 9. Notes, key terms, important questions with answers and MCQs.

Chapter overview

A straight line on the coordinate plane is described by its slope and equation. Different forms of the equation suit different given information.

Slope and forms

Slope m = tan θ = (y₂ − y₁)/(x₂ − x₁); parallel lines m₁ = m₂; perpendicular m₁m₂ = −1.
Point–slope: y − y₁ = m(x − x₁); slope–intercept: y = mx + c; intercept form: x/a + y/b = 1.
General form Ax + By + C = 0 has slope −A/B.

Distances and angles

Angle between lines: tan θ = |(m₂ − m₁)/(1 + m₁m₂)|.
Distance of (x₁, y₁) from Ax + By + C = 0: |Ax₁ + By₁ + C|/√(A² + B²).
Distance between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0: |C₁ − C₂|/√(A² + B²).

Key terms

Slope: tangent of the angle a line makes with the positive x-axis
Intercept: point where a line cuts an axis
Normal form: x cos ω + y sin ω = p, where p is the perpendicular distance from the origin and ω its inclination

Formulas

y − y₁ = m(x − x₁)
d = |Ax₁ + By₁ + C|/√(A² + B²)

Important questions with answers

Find the slope and intercepts of 3x + 4y − 12 = 0.2 marks

Slope = −3/4; x-intercept 4, y-intercept 3.

Find the equation of the line perpendicular to 2x − 3y + 5 = 0 and passing through (1, 2).3 marks

Slope of given line = 2/3, so the required slope = −3/2. y − 2 = −(3/2)(x − 1), i.e. 3x + 2y − 7 = 0.

Find the coordinates of the foot of the perpendicular from (−1, 3) to the line 3x − 4y − 16 = 0, and the length of the perpendicular.5 marks

Length = |3(−1) − 4(3) − 16|/5 = |−31|/5 = 31/5. Foot (h, k): it lies on the line (3h − 4k = 16) and the line from (−1, 3) to (h, k) has slope −4/3 (perpendicular to slope 3/4): (k − 3)/(h + 1) = −4/3, i.e. 4h + 3k = 5. Solving 3h − 4k = 16 and 4h + 3k = 5: h = 68/25, k = −49/25. Foot = (68/25, −49/25).

Quick revision

Parallel ⇒ equal slopes.
Perpendicular ⇒ product −1.
Slope of Ax + By + C = 0 is −A/B.

Exam tips

Three points are collinear if the slopes between pairs are equal.
To find the foot of perpendicular or image of a point in a line, use the slope condition m₁m₂ = −1 together with the mid-point lying on the line.

Practice MCQs

Slope of the line through (1, 2) and (3, 6) is
1. 2
2. 1
3. 4
4. ½
Show answer
Answer: (a) 2. 4/2.
Slope of y = 3x + 5 is
1. 5
2. 3
3. −3
4. 1/3
Show answer
Answer: (b) 3. m in y = mx + c.
Slope of a line perpendicular to a line of slope 2 is
1. 2
2. −½
3. −2
4. ½
Show answer
Answer: (b) −½. m₁m₂ = −1.
Equation of the line with intercepts 2 and 3 on the axes is
1. 3x + 2y = 6
2. 2x + 3y = 6
3. x + y = 5
4. 3x − 2y = 6
Show answer
Answer: (a) 3x + 2y = 6. x/2 + y/3 = 1.
The equation of the line through (2, 3) with slope −1 is
1. x + y = 5
2. x − y = −1
3. x + y = 1
4. y = −x
Show answer
Answer: (a) x + y = 5. y − 3 = −(x − 2).
Distance of (3, 4) from 3x + 4y − 5 = 0 is
1. 4
2. 5
3. 20
4. 25
Show answer
Answer: (a) 4. |9 + 16 − 5|/5.
Distance between 3x + 4y = 5 and 3x + 4y = 15 is
1. 10
2. 1
3. 5
4. 2
Show answer
Answer: (d) 2. 10/5.
The angle between the lines y = √3x + 5 and y = x/√3 − 2 is
1. 30°
2. 45°
3. 60°
4. 90°
Show answer
Answer: (a) 30°. Inclinations 60° and 30°.

All chapters of Class 11 Mathematics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.