Class 11 Maths Chapter 9: Straight Lines

Straight Lines is Chapter 9 of NCERT Class 11 Maths. It builds on Class 10 coordinate geometry and shows how to describe a line using its slope and points on it, how to find the angle between two lines, and how far a point lies from a line. Watch the 6 video lessons below, then use the notes, formulas and solved examples to strengthen your concepts.

What you will learn

A recap of the distance, section and area of triangle formulas
The slope of a line and conditions for parallel and perpendicular lines
The angle between two lines and how to test collinearity
Different forms of the equation of a line
The distance of a point from a line and between two parallel lines

Video Lessons

Straight Lines - Part 1

Straight Lines - Part 2

Straight Lines - Part 3

Straight Lines - Part 4

Straight Lines - Part 5

Straight Lines - Part 6

Key Concepts and Formulas

Coordinate Geometry Recap

These Class 10 results are used again and again in this chapter, so keep them at your fingertips.

Distance = √((x₂ − x₁)² + (y₂ − y₁)²)
Section (internal, m : n):
x = (mx₂ + nx₁)/(m + n)
y = (my₂ + ny₁)/(m + n)
Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Area of triangle =
½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

If the area of the triangle is zero, the three points are collinear

Slope of a Line

The slope tells you how steep a line is. If a line makes an angle θ with the positive direction of the x-axis (measured anticlockwise), θ is called the inclination of the line.

m = tan θ,  θ ≠ 90°
m = (y₂ − y₁)/(x₂ − x₁),  x₁ ≠ x₂

Graph of y = 2x − 3 through (1, −1) and (4, 5) showing slope, Class 11 Straight Lines — Slope = rise ÷ run = 6 ÷ 3 = 2

A line parallel to the x-axis has slope 0; a vertical line has an undefined slope
A positive slope means the line rises from left to right, a negative slope means it falls

Parallel, Perpendicular and Collinear

Slopes give quick tests for how two lines are placed relative to each other.

Parallel: m₁ = m₂
Perpendicular: m₁ × m₂ = −1
Acute angle between two lines:
tan θ = |(m₂ − m₁)/(1 + m₁m₂)|,  1 + m₁m₂ ≠ 0
Collinear A, B, C: slope AB = slope BC

The parallel and perpendicular tests apply to non-vertical lines

Forms of the Equation of a Line

Choose the form that matches the information given in the question. In the end, every line can be written as Ax + By + C = 0.

Horizontal line: y = b
Vertical line: x = a
Point slope: y − y₁ = m(x − x₁)
Two point:
y − y₁ = ((y₂ − y₁)/(x₂ − x₁))(x − x₁)
Slope intercept: y = mx + c
Intercept form: x/a + y/b = 1
Ax + By + C = 0 has slope −A/B  (B ≠ 0)

In y = mx + c, c is the y-intercept; in x/a + y/b = 1, a and b are the x and y intercepts

Distance from a Line

The shortest distance from a point to a line is the length of the perpendicular dropped from the point to the line.

Point (x₁, y₁) to Ax + By + C = 0:
d = |Ax₁ + By₁ + C| / √(A² + B²)
Parallel lines Ax + By + C₁ = 0, Ax + By + C₂ = 0:
d = |C₁ − C₂| / √(A² + B²)

Make sure the coefficients of x and y are the same in both parallel lines before using the second formula

Solved Examples

Example 1: Show that A(1, 2), B(3, 8) and C(5, 14) are collinear.

Slope AB = (8 − 2)/(3 − 1) = 6/2 = 3
Slope BC = (14 − 8)/(5 − 3) = 6/2 = 3
Slopes equal and B is common
∴ A, B and C are collinear

Example 2: Find the acute angle between two lines whose slopes are 2 and 1/3.

tan θ = |(1/3 − 2)/(1 + 2 × 1/3)|
= |(−5/3)/(5/3)|
= 1
∴ θ = 45°

Example 3: Find the equation of the line through (−2, 3) with slope 4.

y − 3 = 4(x + 2)
y = 4x + 8 + 3 = 4x + 11
∴ 4x − y + 11 = 0

Example 4: Find the equation of the line through (1, −1) and (4, 5).

m = (5 − (−1))/(4 − 1) = 6/3 = 2
y + 1 = 2(x − 1)
y = 2x − 3
∴ 2x − y − 3 = 0

Example 5: Find the equation of the line with x-intercept 4 and y-intercept 6.

x/4 + y/6 = 1
Multiply by 12: 3x + 2y = 12
∴ 3x + 2y − 12 = 0

Example 6: Find the distance of the point (2, −3) from the line 3x − 4y + 7 = 0.

d = |3(2) − 4(−3) + 7| / √(3² + 4²)
= |6 + 12 + 7| / √25
= 25/5
∴ d = 5 units

Perpendicular distance from point (2, −3) to the line 3x − 4y + 7 = 0, Class 11 Straight Lines — The perpendicular from P meets the line at (−1, 1)

Example 7: Find the distance between the parallel lines 6x + 8y − 5 = 0 and 6x + 8y + 15 = 0.

C₁ = −5, C₂ = 15, A = 6, B = 8
d = |−5 − 15| / √(36 + 64)
= 20/10
∴ d = 2 units

Common Mistakes to Avoid

Mixing up the order of points in the slope formula. Keep (y₂ − y₁) and (x₂ − x₁) in the same order.
Forgetting the modulus in the angle and distance formulas, which leads to negative answers.
Using the parallel lines distance formula without first making the x and y coefficients identical in both equations.
Saying a vertical line has slope zero. Its slope is undefined; it is the horizontal line that has slope zero.

Quick Revision

m = tan θ = (y₂ − y₁)/(x₂ − x₁)
Parallel: m₁ = m₂
Perpendicular: m₁m₂ = −1
tan θ = |(m₂ − m₁)/(1 + m₁m₂)|
y − y₁ = m(x − x₁)
y = mx + c,  x/a + y/b = 1
Point to line: |Ax₁ + By₁ + C| / √(A² + B²)
Parallel lines: |C₁ − C₂| / √(A² + B²)