A complete guide to CBSE Class 10 Maths, Coordinate Geometry. This page covers all key formulas (Distance, Section, Area, Slope), provides interactive demos to visualize concepts, and includes worked examples and Q&A to help you master the chapter.

A Guide to Coordinate Geometry

CBSE Class 10 Maths: Lines in Two Dimensions

Sort Content

I. The Cartesian Framework

Coordinate geometry connects geometry with algebra. It uses the Cartesian coordinate system to locate points in a plane. This system is defined by two perpendicular lines: the horizontal x-axis and the vertical y-axis. Their intersection is the Origin (0, 0).

Any point P is located by an ordered pair (x, y).

Abscissa (x-coordinate): The perpendicular distance from the y-axis.
Ordinate (y-coordinate): The perpendicular distance from the x-axis.

The axes divide the plane into four Quadrants (I, II, III, IV), each with a specific sign convention. These conventions help you check your work.

Interactive Demo: Quadrant Plotter

Click anywhere on the graph to plot a point. The coordinates and quadrant will update.

Point: (5, 3), Quadrant: I

II. The Distance Formula

A main application of the Cartesian system is calculating the distance between two points, P(x₁, y₁) and Q(x₂, y₂). The formula is a direct use of the Pythagorean theorem.

The Distance Formula

D = √[ (x₂ – x₁)² + (y₂ – y₁)² ]

Using the Formula for Proofs

The Distance Formula is used to verify geometric properties of shapes by calculating side and diagonal lengths. For example, to prove a quadrilateral is a square, you must prove all four sides are equal AND both diagonals are equal.

Quadrilateral Proofs

Shape	Side Lengths	Diagonal Lengths
Parallelogram	Opposite sides are equal.	Diagonals are unequal.
Rectangle	Opposite sides are equal.	Diagonals are equal.
Rhombus	All four sides are equal.	Diagonals are unequal.
Square	All four sides are equal.	Diagonals are equal.

Worked Example: Proving a Square

Prove that A(1, 7), B(4, 2), C(–1, –1), and D(–4, 4) are vertices of a square.

Step 1. Calculate the four side lengths:

AB = √[(4-1)² + (2-7)²] = √[9 + 25] = √34
BC = √[(-1-4)² + (-1-2)²] = √[25 + 9] = √34
CD = √[(-4-(-1))² + (4-(-1))²] = √[9 + 25] = √34
DA = √[(1-(-4))² + (7-4)²] = √[25 + 9] = √34

All sides are equal (√34). The figure is a rhombus.

Step 2. Calculate the two diagonal lengths:

AC = √[(-1-1)² + (-1-7)²] = √[4 + 64] = √68
BD = √[(-4-4)² + (4-2)²] = √[64 + 4] = √68

Both diagonals are equal (√68).

Proof: Because all four sides are equal and both diagonals are equal, the figure ABCD is a square.

III. The Section Formula

The section formula is a tool to find the coordinates of a point P(x, y) that divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in a ratio m : n.

The Internal Section Formula

P(x, y) = ( (mx₂ + nx₁) / (m + n), (my₂ + ny₁) / (m + n) )

The Mid-point Formula

The formulas are connected. The midpoint divides a segment in a 1:1 ratio. If you set m=1 and n=1 in the section formula, it simplifies to the mid-point formula, which is just the average of the coordinates.

The Mid-point Formula (Case of 1:1)

M(x, y) = ( (x₁ + x₂) / 2, (y₁ + y₂) / 2 )

Interactive Demo: Section Formula Slider

Move the slider to change the ratio m:n and see how the point P moves along the line segment from A to B.

Ratio m:n

Ratio: 1:1 | Point P: (5.0, 5.0)

Worked Example: Points of Trisection

Find the two points, P and Q, that divide the segment joining A(4, –1) and B(–2, –3) into three equal parts.

Step 1. Find P (divides AB in ratio 1:2):

P_x = (1(-2) + 2(4)) / (1+2) = 6 / 3 = 2
P_y = (1(-3) + 2(-1)) / (1+2) = -5 / 3
First point is P(2, -5/3).

Step 2. Find Q (divides AB in ratio 2:1):

Q_x = (2(-2) + 1(4)) / (2+1) = 0 / 3 = 0
Q_y = (2(-3) + 1(-1)) / (2+1) = -7 / 3
Second point is Q(0, -7/3).

IV. Advanced Technique: The k:1 Ratio

For problems where you need to find the ratio, it is often much faster to assume the ratio is k : 1 instead of m : n.

This simplifies the section formula, as you only solve for one variable, k.

The k:1 Section Formula

P(x, y) = ( (kx₂ + x₁) / (k + 1), (ky₂ + y₁) / (k + 1) )

Worked Example: Using k:1

Find the ratio in which the point P(–1, 6) divides the segment joining A(–3, 10) and B(6, –8).

Step 1. Assume the ratio is k:1. Use the x-coordinate formula:

x = (k(6) + 1(-3)) / (k + 1)
We know x = -1. So, -1 = (6k – 3) / (k + 1)
-(k + 1) = 6k – 3
–k – 1 = 6k – 3
2 = 7k
k = 2 / 7

Step 2. State the final ratio:

The ratio is k : 1, which is (2/7) : 1.

Multiplying by 7 gives the integer ratio: 2:7.

(You can verify this using the y-coordinate to be certain.)

V. Concept of Slope

The slope (or gradient) of a line is a number that describes its “steepness” and direction. It is the ratio of the “rise” (vertical change) to the “run” (horizontal change).

The Slope Formula

m = (y₂ – y₁) / (x₂ – x₁)

Interpreting the Slope

Positive Slope (m > 0): The line rises from left to right.
Negative Slope (m < 0): The line falls from left to right.
Zero Slope (m = 0): The line is horizontal (e.g., y=5).
Undefined Slope: The line is vertical (e.g., x=2). The formula fails because x₂ – x₁ = 0.

Slope and Collinearity

Slope provides a very fast way to check for collinearity. Three points A, B, and C are collinear if and only if the slope of AB is equal to the slope of BC.

Slope(AB) = Slope(BC) → Points are Collinear

Slope and Perpendicular Lines

Two lines are perpendicular if their slopes are negative reciprocals of each other.

Perpendicular Test

If line 1 has slope m₁ and line 2 has slope m₂, they are perpendicular if:

m₁ × m₂ = -1

(This does not apply to horizontal and vertical lines, which are perpendicular by definition).

VI. Area and Centroid of a Triangle

You can find the area of a triangle given the coordinates of its three vertices: A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃).

The Area of a Triangle Formula

A = ½ | x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) |

Note: The absolute value bars |…| are required because area must be a positive quantity.

Condition for Collinearity

An application of the area formula is to test if three points are collinear (lie on the same straight line). If the points are collinear, they form a “triangle” with an area of zero.

Collinearity Test

If the following expression equals 0, the points are collinear:

x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) = 0

The Centroid of a Triangle

The last concept is the centroid, which is the geometric center of the triangle. It is the point where the three medians (lines from a vertex to the midpoint of the opposite side) intersect. The centroid’s coordinates are simply the average of the vertices’ coordinates.

The Centroid Formula

G(x, y) = ( (x₁ + x₂ + x₃) / 3, (y₁ + y₂ + y₃) / 3 )

Interactive Demo: Triangle Dashboard

Click and drag the vertices A, B, or C to change the triangle’s shape. The Area and Centroid (G) will recalculate live.

Area 15000.0

Centroid (G) (5.0, 3.3)

A Note on History

Coordinate geometry was formalized by French philosopher and mathematician René Descartes in the 17th century.

Legend says he was inspired by watching a fly on his ceiling. He realized he could describe the fly’s path by its distance from two walls (the x and y axes). This linked algebra and geometry, allowing us to “draw” equations and “calculate” shapes.

Deeper Dive: Centroid Property

The Centroid (G) is not just the average of the coordinates. It’s the “center of mass” of the triangle.

A key property is that the centroid divides each median in a 2:1 ratio. The segment from the vertex to the centroid is twice as long as the segment from the centroid to the midpoint of the opposite side.

Problem-Solving Strategy

This guide provides a strategy for solving problems. Use this table to choose the right formula.

Goal	Formula to Use
Find length or prove a shape	Distance Formula
Find a point of division	Section Formula
Find a midpoint	Mid-point Formula
Prove points are on a line	Area Formula (check for 0) or Slope Formula
Find the center of a triangle	Centroid Formula

Quick Q&A

Q1: Find the distance of the point P(2, 3) from the x-axis.

The distance from the x-axis is the definition of the y-coordinate (the ordinate).

The distance is 3. The ordinate (y-coordinate) is the perpendicular distance from the x-axis.

Q2: In what ratio does the y-axis divide the line segment joining P(–4, 5) and Q(3, –7)?

Any point on the y-axis has an x-coordinate of 0. Use the section formula (or k:1 method) for the x-coordinate, set it to 0, and solve for the ratio.

Let the ratio be k:1. The x-coordinate of the dividing point is 0.
x = (k(3) + 1(-4)) / (k + 1) = 0
3k – 4 = 0
k = 4 / 3.
The ratio is (4/3):1, which is 4:3.

Q3: Find the centroid of a triangle whose vertices are (3, –5), (–7, 4), and (10, –2).

The centroid is the average of the x-coordinates and the average of the y-coordinates.

G_x = (3 + (-7) + 10) / 3 = 6 / 3 = 2
G_y = (-5 + 4 + (-2)) / 3 = -3 / 3 = -1
The centroid is G(2, –1).

Frequently Asked Questions

What is the difference between Abscissa and Ordinate?

“Abscissa” is the formal name for the x-coordinate. “Ordinate” is the formal name for the y-coordinate. It is just a way to be specific.

Why does the Area formula have absolute value bars?

Area must be a positive number. Depending on the order you pick the vertices (clockwise vs. anti-clockwise), the formula inside can result in a negative number. The absolute value bars |…| ensure the final answer is always positive, which is correct for a physical area.

Can I use the Distance Formula to check for collinearity?

Yes. Three points A, B, and C are collinear if the sum of the two smaller distances equals the largest distance. For example, if AB + BC = AC. Using the area formula (checking for 0) or the slope formula (checking for equal slopes) is usually much faster and involves less calculation with square roots.

CBSE Coordinate Geometry Class 10: Formulas Examples Demos