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# Find the area of the triangle whose vertices are: (i) (2, 3), (-1, 0), (2, -4) (ii) (-5, -1), (3, -5), (5, 2)

We know that formula for area of a triangle whose vertices are $$(x_1,y_1) , (x_2,y_2) , (x_3,y_3)$$ is,
= $$\frac{1}{2} \ | \ [x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)] \ |$$

(i) So, here $$x_1 \ = \ 2 \ , \ y_1 \ = \ 3 \ , \ x_2 \ = \ -1 \ , \ y_2 \ = \ 0 \ x_3 \ = \ 2 \ , \ y_3 \ = \ -4$$

So, area of triangle = $$\frac{1}{2} \ | \ [ 2( 0 + 4 ) \ - \ 1(-4 - 3) \ + \ 2(3 - 0)] \ | \ = \ \frac{1}{2} \ | \ [ 8 + 7 + 6 ] \ | \ = \ \frac{21}{2}$$

∴ Area of triangle is $$\frac{21}{2}$$ sq. units.

(ii) Similarly, here $$x_1 \ = \ -5 \ , \ y_1 \ = \ -1 \ , \ x_2 \ = \ 3 \ , \ y_2 \ = \ -5 \ x_3 \ = \ 5 \ , \ y_3 \ = \ 2$$
So, area of triangle = $$\frac{1}{2} \ | \ [ (-5)( -5 - 2 ) \ - \ 3(2 + 1) \ + \ 5(-1 + 5)] \ | \ = \ \frac{1}{2} \ | \ [ 35 + 9 + 20 ] \ | \ = \ 32$$

∴ Area of triangle is $$32$$ sq. units.