Class 11 Maths Notes

Straight Lines Class 11 Notes

Complete, exam-ready notes on straight lines: how slope measures steepness, every form of a line equation from point-slope to general, conditions for lines to be parallel or perpendicular, angle between two lines, distance of a point from a line, and area of a triangle formed by three points — written for CBSE boards and JEE revision.

Class11SubjectMathematicsCoversCBSE · JEE

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is this chapter about in one line?

Straight lines connects geometry and algebra: the slope links angle to direction, and every line has an equation you can derive from a point, a slope, or two intercepts.

Slope of a Line

Slope (gradient)

The slope m of a line measures its steepness — the tangent of the angle θ it makes with the positive x-axis: $m = \tan\theta$ . For two points (x₁, y₁) and (x₂, y₂) on the line: $m = \frac{y_2 - y_1}{x_2 - x_1}$ .

A horizontal line has m = 0; a vertical line has m undefined (90° with x-axis).
The angle of inclination θ lies in [0°, 180°); m = tan θ gives a one-to-one mapping.
A positive slope rises left to right; a negative slope falls left to right.

Parallel and Perpendicular Lines

Parallel lines

Two non-vertical lines with slopes m₁ and m₂ are parallel if and only if $m_1 = m_2$ .

Perpendicular lines

Two lines are perpendicular if and only if the product of their slopes is −1: $m_1 \cdot m_2 = -1$ . If one line is vertical, the other must be horizontal.

Exam shortcut

When a question says 'lines making equal angles with axes', it means m₁ and m₂ are negatives of each other — verify using the perpendicular condition.

Forms of the Equation of a Line

Point-slope form: y − y₁ = m(x − x₁) — the workhorse for most problems.
Two-point form: (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁).
Slope-intercept form: y = mx + c, where c is the y-intercept.
Intercept form: x/a + y/b = 1, where a and b are the x- and y-intercepts.
Normal form: x cos α + y sin α = p, where p is the perpendicular distance from origin.

ax + by + c = 0

General form (a, b not both zero)

General Equation and Converting Between Forms

General form

Every line can be written as $ax + by + c = 0$ . From this: slope m = −a/b (b ≠ 0), x-intercept = −c/a (a ≠ 0), y-intercept = −c/b (b ≠ 0). To convert to slope-intercept, solve for y.

Watch the sign

When converting ax + by + c = 0 to y = mx + c form, the slope is −a/b not a/b. This is the most common algebraic slip in board exams.

Angle Between Two Lines

Angle between lines

If two lines have slopes m₁ and m₂, the acute angle θ between them satisfies: $\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|$ . If the denominator is zero (m₁m₂ = −1), the lines are perpendicular and θ = 90°.

The formula gives the acute angle by default; drop the absolute value to get the directed angle.
If one line is vertical (m undefined), θ = |90° − α| where α is the other line's inclination.
For concurrent lines, use the formula pairwise and sum appropriately.

Distance Formulas

Perpendicular distance of a point from a line

The distance from point (x₁, y₁) to line ax + by + c = 0 is: $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$ .

Distance between two parallel lines

For parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0: $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$ . The coefficients a and b must be identical before applying this formula.

Area of a Triangle

Area from three vertices

For vertices (x₁, y₁), (x₂, y₂), (x₃, y₃): $\text{Area} = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ . The area is zero if and only if the three points are collinear.

Collinearity check

To test whether three points are collinear, compute the area — if it equals zero, the points lie on a single straight line.

Solved Examples

Example: Find the equation of the line passing through (2, −3) and making an angle of 135° with the positive x-axis.

Solution: The slope is m = tan 135° = −1. Using point-slope form: y − (−3) = −1(x − 2), which simplifies to x + y + 1 = 0.

Example: Find the distance of the point (3, −5) from the line 4x − 3y − 26 = 0.

Solution: Using the point-to-line distance formula, d = |4(3) − 3(−5) − 26| / √(16 + 9) = |12 + 15 − 26| / 5 = |1| / 5 = 1/5 units.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Slope from angle

m = \tan\theta

Slope from two points

m = \frac{y_2 - y_1}{x_2 - x_1}

Point-slope form

y - y_1 = m(x - x_1)

Slope-intercept form

y = mx + c

Intercept form

\frac{x}{a} + \frac{y}{b} = 1

Distance of a point from a line

d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}

Angle between two lines

\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

m = tan θ always gives the angle with the positive x-axis, measured anticlockwise.
Parallel lines: m₁ = m₂; perpendicular lines: m₁m₂ = −1 — memorise these.
Convert ax + by + c = 0 to slope-intercept to read slope and intercept instantly.
For distance formulas, always reduce the line equation to general form first.
Area formula returning zero is the quickest collinearity test in coordinate geometry.
In angle-between-lines, if m₁m₂ = −1 the denominator is zero — lines are perpendicular.
The intercept form x/a + y/b = 1 directly gives both intercepts; useful for sketching.
Board exams love combining slope + distance — practice converting between forms quickly.

FAQ

Common questions

What is the slope of a line?

The slope m is the tangent of the angle the line makes with the x-axis: m = tan θ. It equals (y₂ − y₁)/(x₂ − x₁) for any two points on the line.

When are two lines parallel?

Two non-vertical lines are parallel if and only if their slopes are equal: m₁ = m₂.

How do you find the distance of a point from a line?

For a line ax + by + c = 0 and point (x₁, y₁), the perpendicular distance is |ax₁ + by₁ + c| / √(a² + b²).

What is the general equation of a straight line?

Any straight line can be written as ax + by + c = 0 where a and b are not both zero. The slope is −a/b when b ≠ 0.

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