FINDING AREA OF A TRIANGLE USING COORDINATES

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 x₁y₂, x₂y₃ and x₃y₁ as shown in the dark arrows. 

Also add the diagonal products x₂y₁, x₃y₂ and x₁y₃ 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)

x₁  =  3     x₂  =  -3      x₃  =  5

y₁  =  9     y₂  =  -9      y₃  =  -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)

x₁  =  4     x₂  =  -2     x₃  =  4

y₁  =  5     y₂  =  2       y₃  =  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)

x₁  =  2       x₂  =  2       x₃  =  3

y₁  =  2       y₂  =  0       y₃  =  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).

x₁  =  0     x₂  =  -3     x₃  =  3

y₁  =  4     y₂  =  1       y₃  =  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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