Coordinate Geometry Class 10: Formulas, Notes & Solved Examples

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Coordinate Geometry Class 10 – What Is CBSE Class 10 Coordinate Geometry?

Coordinate geometry class 10 is the study of geometric figures using algebra, by representing every point on a plane as an ordered pair $(x,y)$. Instead of drawing and measuring with instruments, CBSE Class 10 Coordinate Geometry lets you calculate distances, ratios, and areas purely using coordinates and formulas.

• Every point in coordinate geometry class 10 is written as $(x,y)$, where $x$ is the abscissa and $y$ is the ordinate.
• The chapter builds directly on the Cartesian plane, with axes labelled $x$-axis and $y$-axis, and origin $(0,0)$.
• Coordinate geometry class 10 connects algebra and geometry, and it is a direct prerequisite for coordinate geometry, straight lines, and conic sections in Class 11.

What Is the Distance Formula in Coordinate Geometry Class 10?

The distance formula is the most fundamental result in coordinate geometry class 10, used to find the distance $d$ between any two points $(x_1,y_1)$ and $(x_2,y_2)$ on a plane:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

This formula comes directly from the Pythagoras theorem, treating the horizontal gap $(x_2-x_1)$ and vertical gap $(y_2-y_1)$ between the two points as the two legs of a right triangle, with the distance $d$ as the hypotenuse.

As a special case, the distance of a point $(x,y)$ from the origin $(0,0)$ is given by:

$d = \sqrt{x^2 + y^2}$

What Is the Section Formula in CBSE Class 10 Coordinate Geometry?

The section formula in CBSE Class 10 Coordinate Geometry is used to find the coordinates of a point $P(x,y)$ that divides the line segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ internally in the ratio $ m_1:m_2$:

$x = \frac{m_1x_2 + m_2x_1}{m_1+m_2}$

$y = \frac{m_1y_2 + m_2y_1}{m_1+m_2}$

This is one of the most frequently tested formulas in coordinate geometry class 10 board papers, especially in problems involving trisection of a line segment — dividing it into three equal parts using ratios $1:2$ and $2:1$.

What Is the Midpoint Formula in Coordinate Geometry Class 10?

The midpoint formula is a special case of the section formula, used when a point $P(x,y)$ divides a line segment exactly in half, i.e., in the ratio $1:1$:

$x = \frac{x_1+x_2}{2}$

$y = \frac{y_1+y_2}{2}$

Distance, Section, and Midpoint Formulas at a Glance

What Is the Area of a Triangle Formula in Coordinate Geometry Class 10?

Given three points $A(x_1,y_1)$, $B(x_2,y_2)$, and $C(x_3,y_3)$, the area of the triangle formed by them in coordinate geometry class 10 is:

$\text{Area} = \frac{1}{2}\left|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\right|$

This same formula also gives the collinearity test used throughout CBSE Class 10 Coordinate Geometry — if the calculated area equals $0$, the three points do not form a triangle at all, and are said to be collinear (lying on the same straight line).

How Do You Check if Three Points Are Collinear in Coordinate Geometry Class 10?

To check collinearity of three points $A(x_1,y_1)$, $B(x_2,y_2)$, and $C(x_3,y_3)$ in coordinate geometry class 10, apply the area of triangle formula to them:

• If the area $= 0$, the three points are collinear.
• If the area $\neq 0$, the three points form a proper triangle.

This is a very common short-answer question type in CBSE Class 10 Coordinate Geometry board papers, since it combines the area formula with a simple pass/fail condition.

Solved Examples on Coordinate Geometry Class 10

Example 1: Distance Formula

Find the distance between the points $A(2,3)$ and $B(6,7)$.

$d = \sqrt{(6-2)^2 + (7-3)^2}$

$= \sqrt{16+16}$

$= \sqrt{32}$

$= 4\sqrt{2}$ units

Example 2: Section Formula

Find the coordinates of the point that divides the line segment joining $A(-1,3)$ and $B(4,-2)$ internally in the ratio $2:3$.

$x = \frac{2(4)+3(-1)}{2+3}$

$= \frac{8-3}{5}$

$= 1$

$y = \frac{2(-2)+3(3)}{2+3}$

$= \frac{-4+9}{5}$

$= 1$

The point is $(1,1)$.

Example 3: Midpoint Formula

Find the midpoint of the line segment joining $A(3,-4)$ and $B(7,2)$.

$x = \frac{3+7}{2} = 5$

$y = \frac{-4+2}{2} = -1$

Midpoint $= (5,-1)$.

Example 4: Collinearity Test

Check whether the points $A(4,5)$, $B(7,6)$, and $C(6,3)$ are collinear.

$\text{Area} = \frac{1}{2}\left|4(6-3) + 7(3-5) + 6(5-6)\right|$

$= \frac{1}{2}|12-14-6|$

$= \frac{1}{2}|-8|$

$= 4$

Since the area is $4$ (non-zero), the points are not collinear.

Explore More on eSaral (Related Coordinate Geometry & Class 10 Maths Resources)

Quick revision with formula mind maps: Coordinate Geometry Class 10 Chapter 7 Short Notes (Mind Maps) →
Solve exercise-based questions step by step: NCERT Solutions for Class 10 Maths Chapter 7 — Coordinate Geometry →
Practice Exercise 7.1 in detail: NCERT Solutions Chapter 7 Exercise 7.1 →
Practice Exercise 7.2 in detail: NCERT Solutions Chapter 7 Exercise 7.2 →
Go beyond the textbook with higher-order questions: NCERT Exemplar Class 10 Maths Chapter 7 →
Download the complete syllabus reference: Class 10 Maths NCERT Book PDF →

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