Introduction to Co-Ordinate Geometry

Last Updated : 11 Feb, 2026

Coordinate Geometry is a branch of mathematics that studies the position of points on a plane using numbers called coordinates. It acts as a bridge between algebra and geometry. Coordinate geometry mainly focuses on understanding the coordinate plane, plotting points, and locating points in different quadrants.

The coordinate plane is divided into four quadrants, labeled the first (I), second (II), third (III), and fourth (IV) quadrants.

coordinate

In coordinate geometry, the position of a point is determined by its distances from two fixed perpendicular lines called the X-axis and Y-axis, which together form the Cartesian plane.

Coordinate Axes

The coordinate axes are the two perpendicular lines that form the basis of the Cartesian plane used to represent the position of a point in the plane. To represent a point in the 2D plane, we need 2 coordinate axes, and the plane that is formed by the intersection of these axes is called the Cartesian plane.

These lines are:

1) x-axis:

  • It is the horizontal axis.
  • It represents the horizontal direction, with positive values to the right of the origin and negative values to the left.

2) y-axis:

  • It is the vertical axis.
  • It represents the vertical direction, with positive values above the origin and negative values below.

Cartesian Plane

The Cartesian plane is a two-dimensional plane formed by two perpendicular lines x-axis (horizontal line) and y-axis (vertical line). These axes intersect at a point called the origin (0, 0), which is the center of the plane. The plane is used to locate points using ordered pairs of numbers, called coordinates (x, y), where:

  • x: Represents the position along the x-axis (horizontal direction).
  • y: Represents the position along the y-axis (vertical direction).

The Cartesian plane is divided into four regions, called quadrants, based on the signs of x and y.

Coordinates of a Point

A coordinate of a point is an address, which helps to locate it in space. For a two-dimensional space, the coordinates of a point are (x, y).

The coordinates of a point are useful to perform numerous operations of finding distance, midpoint, the slope of a line, and equation of a line.

coordintes-

Origin: Through a point O, referred to as the origin, we take two mutually perpendicular lines XOX' and YOY', and call them x and y axes respectively.

Abscissa & Ordinate: The position of a point is completely determined with reference to these axes by means of an ordered pair of real numbers (x, y) called the coordinates of P where | x | and | y | are the distances of the point P from the y-axis and the x-axis respectively,

  • Abscissa: It is the x value in the point (x, y), and is the distance of this point along the x-axis, from the origin.
  • Ordinate:  It is the y value in the point (x, y)., and is the perpendicular distance of the point from the x-axis, which is parallel to the y-axis.

Coordinate Geometry Formulas

The formulas of coordinate geometry help in conveniently proving the various properties of lines and figures represented in the coordinate axes.

Distance Formula

The distance between two points A(x1, y1) and B(x2, y2) is given
AB = √((x2 - x1)2 + (y2 - y1)2)

The distance of the point P(x, y) from the origin O(0, 0) is
OP = √((x - 0)2+ (y – 0)2), i.e. OP = √(x² + y²)

Example: Find the distance between the points P(-4, 7) and Q(2, -5).
Solution:

The given points are P(-4, 7) and Q(2, -5).

Then, (x1 = -4, y1 = 7) and (x2 = 2, y2 = -5). : 

PQ =  √((x2 - x1)2 + (y2 - y1)2)
= √((2 - (-4))2 + (-5 - 7)2)
= √(62 + (-12)2
= √(36 + 144)
= √180 
= 6√5 

Mid-Point Formula

The midpoint is the point that divides any line segment into two equal parts. The formula to find the midpoint of the line with endpoints A(x1, y1) and B(x2, y2) is given by,

midpoint (x ,y) = {(x1 + x2)/2 , (y1 + y2)/2}

mid_point_of_line_

Example: Find the coordinates of the midpoint of the line segment joining points A(-5, 4) and B(7,-8).

Solution:

Let M(x, y) be the midpoint of AB. Then, 
x = ((-5) + 7)/2 = 1 and  y = ((4 + (-8))/2 = -2

Hence, the required point is M (1, -2).

Section Formula

The coordinates of the point P(x, y) which divides the line segment joining A(x1, y1) and B(x2, y2) internally in the ratio m:n are given by 

x = (mx2 + nx1) / (m + n)

y = (my2 + ny1) / (m + n)

Example: Find the coordinates of the point that divides the line segment joining the points A(5, 4) and B(7,8) in the ratio of 2:3.

Solution:

Let M(x, y) be the required point,
given m:n = 2:3 and (x1, y1) = (5, 4) , (x2, y2) = (7,8)

now for M(x, y)

x = (mx2 + nx1) / (m + n)
= (2 × 7 + 3 × 5)/(2 + 3)
= 29/5

y = (my2 + ny1) / (m + n)
= (2 × 8 + 3 × 4)/(2 + 3)
= 28/5

So, the required point M(x, y) is (29/5 , 28/5)

Slope Formula

The slope of a line is the inclination of the line. Slope Formula is calculated by measuring the angle made by the line with the positive x-axis, or by calculating it using any two points on the line. The slope of a line inclined at an angle θ with the positive x-axis is given by m = tan θ.  The slope of a line joining the two points A(x1, y1) and B(x2, y2) is given by

m = tan θ = (y2-y1) / (x2-x1)

Example: Find the slope of the line segment joining by points A(5, 1) and B(7,8).

Solution:

given,  (x1, y1) = (5, 1) , (x2, y2) = (7, 8)

Slope (m) = (y2-y1) / (x2-x1)
= (8-1)/(7-5)
= 7/2

Area of Triangle

The area of a triangle ABC with vertices A(x1, y1), B(x2, y2), and C(x3, y3) is given by 

area(ABC) = |1/2 {x1(y2 - y3) + x2(y3 – y1) + x3 (y1 - y2)}|

Example: Find the area of the triangle whose vertices are A(2, 7), B(3, -1), and C(-5, 6).

Solution:

Let A(2, 7), B(3, -1) and C(-5, 6) be the vertices of the given â–³ABC.

Then,
(x1 = 2, y1 = 7), (x2 = 3, y2 = -1) and (x3 = -5, y3 = 6). 

Area of △ABC =  |1/2 {x1(y2 - y3) + x2(y3 – y1) + x3 (y1 - y2)}|

= 1/2 |2(-1 - 6) + 3(6 - 7) – 5(7 +1)|
= 1/2| -14 - 3 - 40|
= 1/2|-57|
= 57/2
= 28.5 sq units.

Condition for Collinearity of Three Points

Let the given points be A(x1, y1), B(x2, y2), and C(x3, y3). Then A, B, and C are collinear,

 area of ABC = 0
1/2[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] = 0

OR

Slope of AB = Slope of BC

Example: Show that points A(-1, 1), B(5, 7), and C(8,10) are collinear.

Solution: 

Let A(-1, 1), B(5, 7) and C(8, 10) be the given points. 

Then, (x1 = -1, y1 = 1), (x2 = 5, y2 = 7) and (x3 = 8, y3 = 10)

∴  x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)

= (-1) (7 – 10) + 5(10 -1) + 8(1 – 7) 
= (3 + 45 - 48)
= 0 

Hence, the given points are collinear. 

Centroid of a Triangle

Centroid of a triangle is the point of intersection of all three medians of a triangle. The centroid of a triangle having its vertices A(x1,y1), B(x2,y2), and C(x3,y3) is given by the formula

Centroid (x, y) = {(x1 + x2 + x3)/3, (y1 + y2 + y3)/3}

Example: Find the centroid of the triangle whose vertices are A(2, 7), B(3, -1), and C(-5, 6).

Solution:

Let A(2, 7), B(3, -1) and C(-5, 6) be the vertices of the given â–³ABC.

(x1 = 2, y1 = 7), (x2 = 3, y2 = -1) and (x3 = -5, y3 = 6).

Centroid (x, y) = {(x1 + x2 + x3)/3 , (y1 + y2 + y3)/3}

= {2 + 3 + ( -5)}/3 , {7 + (-1) + 6}/3

= (2 + 3 - 5)/3 = 0 , y = (7 - 1 + 6)/3 = 12/3 = 4
= (0 , 4)

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Unsolved Question on Co - Ordinates Geometry

Question 1: Find the distance between the points A(3, −4) and B(7, 2).

Question 2: The midpoint of AB is (3, −1). If A is (1, 2), find the coordinates of B.

Question 3: The vertices of a triangle are (−2, 4), (4, −2), and (6, 8). Find its centroid.

Question 4: Show that the points (2, −2), (4, −4), and (6, −6) lie on a straight line.

Question 5: Find the area of the triangle formed by the points (1, −2), (−3, 4), and (5, 6).

Question 6: Find the slope of the line joining (6, −2) and (6, 4).

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