## FINDING AREA OF A TRIANGLE USING COORDINATES

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 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 ## Find Area of a Triangle Using Coordinates Examples

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      y =  -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       y =  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       y =  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       y =  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. After having gone through the stuff given above, we hope that the students would have understood how to find the area of the triangle with the given three vertices.

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