Class 10 Maths Chapter 7: Coordinate Geometry

Coordinate Geometry is Chapter 7 of NCERT Class 10 Maths. It shows how to measure the distance between two points and how to find a point that divides a line segment in a given ratio, using only their coordinates. Watch the video lesson below, then use these notes, formulas and solved examples to practise.

What you will learn

A quick recap of the Cartesian plane, axes, quadrants and coordinates
The distance formula and the distance of a point from the origin
Using distances to check whether points are collinear or form a particular figure
The section formula for internal division of a line segment
The mid-point formula and finding points of trisection

Video Lessons

Coordinate Geometry - Part 1

Key Concepts and Formulas

The Cartesian Plane

Every point in a plane can be located by an ordered pair (x, y). The x-coordinate (abscissa) gives the distance from the y-axis, and the y-coordinate (ordinate) gives the distance from the x-axis, with signs showing the direction.

Signs in the quadrants: I (+, +), II (−, +), III (−, −), IV (+, −)
Any point on the x-axis is (x, 0) and any point on the y-axis is (0, y)
The origin is O(0, 0)

Distance Formula

The distance formula comes straight from the Pythagoras relation applied to a right triangle whose legs are the horizontal and vertical gaps between the two points.

PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]
Distance from origin: OP = √(x² + y²)

Points A(2, −3) and B(8, 5) on a grid with a right triangle explaining the distance formula — Gaps of 6 and 8 give a distance of 10

Distance is never negative, and the order of the points does not change the answer

Applications of the Distance Formula

Many CBSE board exam questions ask you to identify a figure or test points using only distances.

Collinear A, B, C: AB + BC = AC
Right triangle: (longest side)² = sum of squares of others

Isosceles triangle: two sides equal; equilateral triangle: all three sides equal
Parallelogram: opposite sides equal; rhombus: all four sides equal; rectangle: opposite sides equal and diagonals equal; square: all sides equal and diagonals equal

Section Formula

If a point P divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂, its coordinates can be written directly.

x = (m₁x₂ + m₂x₁)/(m₁ + m₂)
y = (m₁y₂ + m₂y₁)/(m₁ + m₂)

Section formula on a grid: P(2, −1) divides A(−2, 3) to B(8, −7) in ratio 2 : 3 — P divides AB internally in the ratio 2 : 3

When the ratio is unknown, take it as k : 1 and solve for k
A point on the x-axis has y = 0, and a point on the y-axis has x = 0; use this to find the ratio in which an axis divides a segment

Mid-point and Trisection

The mid-point is just the section formula with ratio 1 : 1. Points of trisection divide a segment into three equal parts.

Mid-point = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Trisection points: ratios 1 : 2 and 2 : 1

Diagonals of a parallelogram bisect each other, so both diagonals share the same mid-point

Solved Examples

Example 1: Find the distance between A(2, −3) and B(8, 5).

AB = √[(8 − 2)² + (5 − (−3))²]
AB = √(6² + 8²)
AB = √(36 + 64) = √100
∴ AB = 10 units

Example 2: Find x if the distance between P(x, 4) and Q(1, 1) is 5 units.

(x − 1)² + (4 − 1)² = 5²
(x − 1)² + 9 = 25
(x − 1)² = 16
x − 1 = ±4
∴ x = 5 or x = −3

Example 3: Check whether A(1, 2), B(3, 6) and C(4, 8) are collinear.

AB = √(2² + 4²) = √20 = 2√5
BC = √(1² + 2²) = √5
AC = √(3² + 6²) = √45 = 3√5
AB + BC = 2√5 + √5 = 3√5 = AC
∴ A, B, C are collinear

Example 4: Find the point that divides the segment joining (−2, 3) and (8, −7) internally in the ratio 2 : 3.

m₁ = 2, m₂ = 3
x = (2 × 8 + 3 × (−2))/5 = (16 − 6)/5 = 2
y = (2 × (−7) + 3 × 3)/5 = (−14 + 9)/5 = −1
∴ P = (2, −1)

Example 5: In what ratio does the x-axis divide the segment joining (3, −4) and (6, 8)? Also find the point.

Let ratio = k : 1
y = (8k − 4)/(k + 1) = 0
8k = 4, so k = 1/2 → ratio 1 : 2
x = (1 × 6 + 2 × 3)/3 = 12/3 = 4
∴ Ratio 1 : 2, point (4, 0)

Example 6: Three vertices of parallelogram ABCD are A(1, 2), B(4, 3) and C(6, 6). Find D.

Let D = (x, y)
Mid-point of AC = ((1 + 6)/2, (2 + 6)/2) = (3.5, 4)
Mid-point of BD = ((4 + x)/2, (3 + y)/2)
(4 + x)/2 = 3.5, so x = 3
(3 + y)/2 = 4, so y = 5
∴ D = (3, 5)

Example 7: Find the points of trisection of the segment joining (0, 3) and (9, −3).

Ratio 1 : 2:
x = (1 × 9 + 2 × 0)/3 = 3
y = (1 × (−3) + 2 × 3)/3 = 1
Ratio 2 : 1:
x = (2 × 9 + 1 × 0)/3 = 6
y = (2 × (−3) + 1 × 3)/3 = −1
∴ Points are (3, 1) and (6, −1)

Common Mistakes to Avoid

Swapping m₁ and m₂ in the section formula; m₁ is always multiplied by the coordinates of the second point B.
Sign errors with negative coordinates, for example writing (5 − (−3)) as 2 instead of 8.
Forgetting the ± when taking a square root, which loses one of the two possible answers.
To prove a figure is a square or rectangle, showing only that the sides are equal; you must also show the diagonals are equal.

Quick Revision

Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]
From origin = √(x² + y²)
Section (m₁ : m₂):
x = (m₁x₂ + m₂x₁)/(m₁ + m₂)
y = (m₁y₂ + m₂y₁)/(m₁ + m₂)
Mid-point = ((x₁ + x₂)/2, (y₁ + y₂)/2)