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Mathematics Grade XI

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Chapter 9: Straight Lines

Class 11 • Mathematics • Chapter 9

Straight Lines

Describing a straight line with a single equation, and reading its slope and position from it.

CBSE / NCERT Aligned

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Chapter Roadmap

Slope of a Line • Parallel and Perpendicular • Point-Slope and Two-Point • Slope-Intercept and Intercept • General and Normal Form • Distance of a Point • Key Results • Applications

Why Straight Lines Matter

A straight line is the simplest shape in coordinate geometry, yet it sits behind a huge amount of mathematics and science. A road on a map, the path of light, the relationship between cost and quantity, and the trend line through a set of readings can all be modelled by straight lines. The great idea of coordinate geometry is that every straight line can be captured by a single equation, so geometry questions become algebra questions that we can solve neatly.

In this chapter you will learn to measure how steep a line is using its slope, to tell when two lines are parallel or at right angles, to write the equation of a line in several handy forms, and to find the shortest distance from a point to a line. Each form of the equation is suited to a particular piece of information, so a large part of the skill is choosing the right form for the question in front of you.

Key idea

The slope m tells you how steep a line is and which way it tilts. Two lines are parallel when their slopes are equal, and perpendicular when the product of their slopes is −1. Once you know a slope and one point, the whole line is fixed.

Key Terms You Must Know

Term

Meaning

Example

Slope

A measure of steepness, m = (change in y)/(change in x).

A line through (0,0) and (1,2) has slope 2

Intercept

Where a line cuts an axis; the x-intercept and y-intercept.

y = 2x + 3 has y-intercept 3

Point-slope form

y − y₁ = m(x − x₁), using one point and the slope.

Through (1,2) with m = 3

Slope-intercept form

y = mx + c, using the slope and the y-intercept.

y = 2x − 5

Intercept form

x/a + y/b = 1, where a and b are the intercepts.

x/3 + y/2 = 1

General form

Ax + By + C = 0, with slope −A/B.

3x + 4y − 12 = 0

Distance from a line

Shortest (perpendicular) distance from a point to a line.

|Ax₁ + By₁ + C| / √(A² + B²)

Core Concepts, Step by Step

1. The Slope of a Line

The slope (or gradient) measures how steep a line is. For two points (x₁, y₁) and (x₂, y₂) on the line, the slope is m = (y₂ − y₁)/(x₂ − x₁), the change in y divided by the change in x. If the line makes an angle θ with the positive x-axis, then m = tan θ. A line rising to the right has a positive slope, one falling to the right has a negative slope, a horizontal line has slope 0, and a vertical line has no defined slope.

Slope: rise divided by run

Line on coordinate axes showing slope as rise over run

2. Parallel and Perpendicular Lines

Slopes make it very easy to compare directions. Two lines are parallel exactly when their slopes are equal, m₁ = m₂, because they tilt by the same amount. Two lines are perpendicular exactly when the product of their slopes is −1, that is m₁ m₂ = −1. More generally, the angle θ between two lines with slopes m₁ and m₂ satisfies tan θ = |(m₁ − m₂)/(1 + m₁ m₂)|. Three points are collinear (lie on one line) when the slope between each pair is the same.

3. The Point-Slope and Two-Point Forms

If you know one point (x₁, y₁) on a line and its slope m, the equation is y − y₁ = m(x − x₁); this is the point-slope form. If instead you know two points, first work out the slope from them, then use either point in the point-slope form. This pair of forms covers the most common situations, where a line is fixed by a point and a direction, or by two points it passes through.

4. The Slope-Intercept and Intercept Forms

When you know the slope m and the y-intercept c (where the line crosses the y-axis), the equation is simply y = mx + c, the slope-intercept form. When you know the x-intercept a and the y-intercept b, the equation is x/a + y/b = 1, the intercept form. The intercept form is very quick for sketching a line, because you just mark a on the x-axis and b on the y-axis and join them. The table below gathers all the forms together.

The standard forms of the equation of a line

Form

Equation

Use it when you know

Point-slope

y − y₁ = m(x − x₁)

one point and the slope

Two-point

slope first, then point-slope

two points on the line

Slope-intercept

y = mx + c

the slope and the y-intercept

Intercept

x/a + y/b = 1

both axis intercepts

General

Ax + By + C = 0

slope is −A/B

5. The General Form and the Normal Form

Any straight line can be written in the general form Ax + By + C = 0. From it the slope is read off as −A/B (when B is not 0). This single form contains all the others. The normal form x cos ω + y sin ω = p describes the line by the length p of the perpendicular dropped to it from the origin and the angle ω that perpendicular makes with the x-axis. It is useful whenever a problem talks about the perpendicular distance from the origin.

6. Distance of a Point From a Line

The shortest distance from a point to a line is measured along the perpendicular. For the point (x₁, y₁) and the line Ax + By + C = 0, this distance is d = |Ax₁ + By₁ + C| / √(A² + B²). The absolute value bars keep the distance positive whichever side of the line the point is on. For two parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0, the distance between them is |C₁ − C₂| / √(A² + B²).

Key Results & Proofs

Three results underpin the whole chapter. Each follows quickly from the meaning of slope.

Result 1: The Point-Slope Form

Statement. The line through the point (x₁, y₁) with slope m has equation y − y₁ = m(x − x₁).

Proof

Use the fact that the slope between any two points of a line is the same.

Let (x, y) be any other point on the line. – take a general point

The slope from (x₁, y₁) to (x, y) must equal m. – slope is constant on a line

(y − y₁)/(x − x₁)

=

m

write the slope between the two points

y − y₁

=

m(x − x₁)

multiply both sides by (x − x₁)

Putting (x₁, y₁) = (0, c) gives the slope-intercept form y = mx + c at once.

Result 2: The Perpendicular Condition

Statement. Two lines with slopes m₁ and m₂ are perpendicular if and only if m₁ m₂ = −1.

Proof

Compare the angles the two lines make with the x-axis.

Let the lines make angles θ₁ and θ₂ with the x-axis, so m₁ = tan θ₁ and m₂ = tan θ₂. – slope is the tangent of the angle

If the lines are perpendicular, then θ₂ = θ₁ + 90°. – right angle between them

tan θ₂

=

tan(θ₁ + 90°) = −cot θ₁

shift the angle by 90 degrees

m₂

=

−1/m₁

since cot = 1/tan

m₁ m₂

=

−1

multiply across

So a line of slope 2 is perpendicular to a line of slope −1/2, because 2 × (−1/2) = −1.

Result 3: The Slope of the General Form

Statement. The line Ax + By + C = 0 (with B not 0) has slope −A/B.

Proof

Rearrange the general form into slope-intercept form.

Ax + By + C

=

0

general form

By

=

−Ax − C

move the other terms across

y

=

(−A/B)x − C/B

divide by B

Comparing with y = mx + c, the slope is the coefficient of x. – read off the slope

m

=

−A/B

the result

So 3x + 4y − 12 = 0 has slope −3/4, and its y-intercept is 12/4 = 3.

Worked Examples

Example 1

Question: Find the slope of the line through the points (2, 3) and (5, 9).

▶ Show full working

Use m = (y₂ − y₁)/(x₂ − x₁).

m

=

(9 − 3)/(5 − 2)

substitute the points

m

=

6/3 = 2

simplify

Answer: The slope is 2.

Example 2

Question: Find the slope of the line 3x + 4y = 12.

▶ Show full working

Write it as 3x + 4y − 12 = 0 and use slope = −A/B.

m

=

−A/B

slope of the general form

m

=

−3/4

with A = 3 and B = 4

Answer: The slope is −3/4.

Example 3

Question: Are the lines y = 2x + 1 and y = 2x − 5 parallel?

▶ Show full working

Compare their slopes.

Both lines are in slope-intercept form y = mx + c. – read off the slopes

Each has slope 2, and the slopes are equal. – compare

Answer: Yes, they are parallel.

Example 4

Question: Are two lines with slopes 3 and −1/3 perpendicular?

▶ Show full working

Multiply the slopes and check for −1.

m₁ m₂

=

3 × (−1/3)

multiply the slopes

m₁ m₂

=

−1

the perpendicular condition holds

Answer: Yes, the lines are perpendicular.

Example 5

Question: Find the equation of the line through (1, 2) with slope 3.

▶ Show full working

Use the point-slope form y − y₁ = m(x − x₁).

y − 2

=

3(x − 1)

substitute the point and slope

y − 2

=

3x − 3

open the bracket

y

=

3x − 1

make y the subject

Answer: y = 3x − 1.

Example 6

Question: Find the equation of the line through (2, 3) and (4, 7).

▶ Show full working

Find the slope first, then use the point-slope form.

m

=

(7 − 3)/(4 − 2) = 2

slope from the two points

y − 3

=

2(x − 2)

point-slope form with (2, 3)

y

=

2x − 1

simplify

Answer: y = 2x − 1.

Example 7

Question: Write the equation of the line with slope 2 and y-intercept −5.

▶ Show full working

Use the slope-intercept form y = mx + c directly.

y

=

mx + c

slope-intercept form

y

=

2x − 5

substitute m = 2 and c = −5

Answer: y = 2x − 5.

Example 8

Question: Write the line 2x + 3y = 6 in intercept form and state its intercepts.

▶ Show full working

Divide through so the right side becomes 1.

2x + 3y

=

6

given

x/3 + y/2

=

1

divide both sides by 6

Comparing with x/a + y/b = 1 gives a = 3, b = 2. – read off the intercepts

Answer: Intercept form x/3 + y/2 = 1; x-intercept 3 and y-intercept 2.

Example 9

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

▶ Show full working

Use d = |Ax₁ + By₁ + C| / √(A² + B²).

d

=

|3(3) + 4(4) − 5| / √(3² + 4²)

substitute the point

d

=

|9 + 16 − 5| / √25

work out top and bottom

d

=

20 / 5 = 4

simplify

Answer: The distance is 4 units.

Example 10

Question: Find the distance between the parallel lines 3x + 4y + 5 = 0 and 3x + 4y − 10 = 0.

▶ Show full working

Use d = |C₁ − C₂| / √(A² + B²).

d

=

|5 − (−10)| / √(3² + 4²)

substitute C₁ = 5, C₂ = −10

d

=

15 / 5

simplify

d

=

3

the distance

Answer: The distance is 3 units.

Example 11

Question: Find the equation of the line through the origin and the point (3, 6).

▶ Show full working

The origin is (0, 0); find the slope, then use slope-intercept form with c = 0.

m

=

(6 − 0)/(3 − 0) = 2

slope from the two points

y

=

2x

through the origin, so c = 0

Answer: y = 2x.

Example 12

Question: Find k if the lines 2x + ky = 5 and 3x − 2y = 7 are perpendicular.

▶ Show full working

Find each slope using −A/B, then use m₁ m₂ = −1.

m₁

=

−2/k

slope of 2x + ky = 5

m₂

=

3/2

slope of 3x − 2y = 7

(−2/k)(3/2)

=

−1

perpendicular condition

−3/k

=

−1

simplify the left side

k

=

3

solve

Answer: k = 3.

Where You Meet This in Real Life

Maps and navigation

Roads, flight paths and the grid of a city are modelled with straight lines, and slope tells you how steep a climb or descent is.

Economics and business

Cost, revenue and demand often vary linearly with quantity, so their graphs are straight lines whose slope carries real meaning.

Science and readings

When two measured quantities are related linearly, the trend line through the readings lets you predict values you did not measure.

Construction and design

Ramps, staircases and roofs are built to set gradients, which are exactly the slopes of straight lines.

Computer graphics

Drawing and moving shapes on a screen relies on the equations of lines and the distances between points and lines.

Practice Sets A to D

Practice Set A – Basics

A1. Find the slope of the line through (1, 1) and (4, 7).

▶ Reveal full working

Use m = (y₂ − y₁)/(x₂ − x₁).

m

=

(7 − 1)/(4 − 1) = 6/3

substitute

m

=

2

simplify

Answer: 2.

A2. Find the slope of the line 2x − y + 3 = 0.

▶ Reveal full working

Use slope = −A/B.

m

=

−A/B = −2/(−1)

with A = 2, B = −1

m

=

2

simplify

Answer: 2.

A3. Write the equation of the line through the origin with slope −2.

▶ Reveal full working

Use slope-intercept form with c = 0.

y

=

−2x

slope −2 through the origin

Answer: y = −2x.

A4. State the y-intercept of the line y = 3x + 7.

▶ Reveal full working

Compare with y = mx + c.

In y = mx + c, the y-intercept is c.

Here c = 7.

Answer: 7.

Practice Set B – Conceptual

B1. What is the condition for two lines to be parallel?

▶ Reveal full working

Compare their slopes.

Parallel lines tilt by the same amount.

So their slopes are equal: m₁ = m₂.

Answer: Their slopes are equal.

B2. What is the condition for two lines to be perpendicular?

▶ Reveal full working

Look at the product of the slopes.

Perpendicular lines meet at a right angle.

The product of their slopes is −1: m₁ m₂ = −1.

Answer: The product of their slopes is −1.

B3. What is the slope of a horizontal line, and of a vertical line?

▶ Reveal full working

Think about the change in y and the change in x.

A horizontal line has no change in y, so its slope is 0.

A vertical line has no change in x, so its slope is undefined.

Answer: Horizontal: slope 0; vertical: slope undefined.

B4. What do a and b mean in the intercept form x/a + y/b = 1?

▶ Reveal full working

Set one variable to 0 at a time.

Putting y = 0 gives x = a, the x-intercept.

Putting x = 0 gives y = b, the y-intercept.

Answer: a is the x-intercept and b is the y-intercept.

Practice Set C – Application / Numerical

C1. Find the equation of the line through (2, −1) with slope 4.

▶ Reveal full working

Use the point-slope form.

y − (−1)

=

4(x − 2)

substitute

y + 1

=

4x − 8

open the bracket

y

=

4x − 9

make y the subject

Answer: y = 4x − 9.

C2. Find the equation of the line through (1, 2) and (3, 8).

▶ Reveal full working

Find the slope, then use point-slope form.

m

=

(8 − 2)/(3 − 1) = 3

slope

y − 2

=

3(x − 1)

point-slope form

y

=

3x − 1

simplify

Answer: y = 3x − 1.

C3. Find the distance of the point (1, 2) from the line 4x + 3y − 10 = 0.

▶ Reveal full working

Use the distance formula.

d

=

|4(1) + 3(2) − 10| / √(16 + 9)

substitute

d

=

|4 + 6 − 10| / 5

simplify

d

=

0 / 5 = 0

the point lies on the line

Answer: 0 (the point is on the line).

C4. Find the distance between the parallel lines 5x − 12y + 6 = 0 and 5x − 12y − 7 = 0.

▶ Reveal full working

Use d = |C₁ − C₂| / √(A² + B²).

d

=

|6 − (−7)| / √(25 + 144)

substitute

d

=

13 / 13

since √169 = 13

d

=

1

the distance

Answer: 1 unit.

Practice Set D – HOTS / Word Problems

D1. Show that the points (1, 4), (3, −2) and (−3, 16) are collinear.

▶ Reveal full working

Three points are collinear when the slope between each pair is the same.

slope of first pair

=

(−2 − 4)/(3 − 1) = −3

from (1,4) to (3,−2)

slope of second pair

=

(16 − (−2))/(−3 − 3) = −3

from (3,−2) to (−3,16)

The two slopes are equal, so the points lie on one line. – conclusion

Answer: They are collinear (each slope is −3).

D2. A line makes intercepts 3 and 4 on the x-axis and y-axis. Find its equation and its distance from the origin.

▶ Reveal full working

Use the intercept form, then the distance formula from the origin.

x/3 + y/4

=

1

intercept form

4x + 3y − 12

=

0

multiply by 12 and rearrange

d

=

|4(0) + 3(0) − 12| / √(16 + 9)

distance from the origin

d

=

12 / 5 = 2.4

simplify

Answer: Equation 4x + 3y − 12 = 0; distance 2.4 units.

D3. Find k if the line through (2, k) and (3, 5) has slope 2.

▶ Reveal full working

Set the slope between the points equal to 2.

(5 − k)/(3 − 2)

=

2

slope between the points

5 − k

=

2

simplify the denominator

k

=

3

solve

Answer: k = 3.

D4. Find the distance between the parallel lines 2x + 3y = 7 and 2x + 3y = 14.

▶ Reveal full working

Write both as Ax + By + C = 0, then use the parallel-line distance formula.

The lines are 2x + 3y − 7 = 0 and 2x + 3y − 14 = 0. – rewrite in general form

d

=

|−7 − (−14)| / √(4 + 9)

apply the formula

d

=

7 / √13

simplify

Answer: 7/√13 units (about 1.94).

Chapter Summary

Everything in One Glance

Slope

m = (change in y)/(change in x) = tan θ. Positive rises, negative falls, horizontal is 0, vertical undefined.

Parallel and Perpendicular

Parallel: equal slopes. Perpendicular: product of slopes is −1.

Point and Two-Point Forms

y − y₁ = m(x − x₁); for two points, find the slope first.

Slope-Intercept and Intercept

y = mx + c uses slope and y-intercept; x/a + y/b = 1 uses the two intercepts.

General Form

Ax + By + C = 0 has slope −A/B and contains all the other forms.

Distance

Point to line: |Ax₁ + By₁ + C|/√(A² + B²). Between parallels: |C₁ − C₂|/√(A² + B²).

Are You Exam-Ready?

8-Point Exam Quick-Check

Find the slope of the line through (0, 0) and (2, 6).

Find the slope of the line 4x − 2y + 1 = 0.

Are y = 3x + 2 and y = 3x − 7 parallel?

Are the slopes 5 and −1/5 perpendicular?

Write the equation of the line through (2, 5) with slope −1.

Write 3x + 4y = 24 in intercept form.

Find the distance of (0, 0) from 3x + 4y − 10 = 0.

What is the slope of the line x = 7?

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Class 11 Mathematics Chapter 9: Straight Lines, Complete Notes and Practice

These free Class 11 Maths Chapter 9 Straight Lines notes follow the latest NCERT 2026-27 syllabus and give complete, exam-ready coverage for board exams and competitive foundations. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.