In Geometry, a triangle is the 3 – sided polygon which has 3 edges and 3 vertices.

Area of the triangle is a measure of the space covered by the triangle in the two-dimensional plane.

In this article, you will learn how to find the area of a triangle in the coordinate geometry.

**Finding Area of a Triangle Using Coordinates :**

When we have vertices of the triangle and we need to find the area of the triangle, we can use the following steps.

(i) Plot the points in a rough diagram.

(ii) Take the vertices in counter clock-wise direction. Otherwise the formula gives a negative value.

(iii) Use the formula given below

And the diagonal products x_{1}y_{2}, x_{2}y_{3} and x_{3}y_{1} as shown in the dark arrows.

Also add the diagonal products x_{2}y_{1}, x_{3}y_{2} and x_{1}y_{3} as shown in the dotted arrows.

Now, subtract the latter product from the former product to get area of the triangle ABC.

So, area of the triangle ABC is

**Example 1 :**

Find the area of triangle whose vertices are (-3,-9) (-1,6) and (3,9).

**Solution :**

First we have to plot the point in the graph sheet as below.

Now we have to take anticlockwise direction. So we have to take the points in the order C (3, 9) B (-1, 6) and A (-3, -9)

Area of the triangle CBA

= (1/2) {(18 + 9 - 18) - (-9 - 18 - 27)}

= (1/2) {9 - ( -54)}

= (1/2) {9 + 54}

= (1/2) (63)

= (63/2)

= 31.5 Square units.

Therefore the area of CBA = 31.5 square units.

**Example 2 :**

Find the area of triangle whose vertices are (-3, -9) (3, 9) and (5, -8).

**Solution :**

First we have to plot the point in the graph sheet as below.

Now we have to take anticlockwise direction. So we have to take the points in the order B (3, 9) A (-3, -9) and C (5, -8)

x_{1} = 3 x_{2} = -3 x_{3} = 5

y_{1} = 9 y_{2} = -9 y_{3 } = -8

Area of the triangle BAC

= (1/2) {(-27 - 24 + 45) - (-27 - 45 - 24)}

= (1/2) {(-51 + 45) - (-96)}

= (1/2) {-6 + 96}

= (1/2) (90)

= (90/2)

= 45 Square units.

Therefore the area of BAC = 45 square units.

**Example 3 :**

Find the area of triangle whose vertices are (4, 5) (4, 2) and (-2, 2).

**Solution :**

Now we have to take anticlockwise direction. So we have to take the points in the order A (4, 5) C (-2, 2) and B (4, 2)

x_{1} = 4 x_{2} = -2 x_{3 }= 4

y_{1} = 5 y_{2} = 2 y_{3 } = 2

Area of the triangle ACB

= (1/2) {(8 - 4 + 20) - (-10 + 8 + 8)}

= (1/2) {(28 - 4) - ( -10 + 16)}

= (1/2) (24 - 6)

= (1/2) x 18

= (18/2)

= 9 Square units.

Therefore the area of ACB = 9 square units

**Example 4 :**

Find the area of triangle whose vertices are (3, 1) (2, 2) and (2, 0).

**Solution :**

Now we have to take anticlockwise direction. So we have to take the points in the order B (2, 2) C (2, 0) and A (3, 1)

x_{1} = 2 x_{2} = 2 x_{3} = 3

y_{1} = 2 y_{2 } = 0 y_{3} = 1

Area of the triangle BCA

= (1/2) {(0 + 2 + 6) - (4 + 0 + 2)}

= (1/2) {8 - 6}

= (1/2) (2)

= (2/2)

= 1 Square units.

Therefore the area of ACB = 1 square units.

**Example 5 :**

Find the area of triangle whose vertices are (3, 1) (0, 4) and (-3, 1).

**Solution :**

Now we have to take anticlockwise direction. So we have to take the points in the order B (0, 4) C (-3, 1) and A (3, 1).

x_{1} = 0 x_{2} = -3 x_{3} = 3

y_{1} = 4 y_{2} = 1 y_{3 } = 1

Area of the triangle BCA

= (1/2) {(0 -3 + 12) - (-12 + 3 + 0)}

= (1/2) {9 - ( -12 + 3)}

= (1/2) {9 - (-9) }

= (1/2) {9 + 9)}

= (1/2) x 18

= 9 Square units.

Therefore the area of BCA = 9 square units.

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