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Find the area of triangle whose vertices is A(2,3),B(-2,1)and C(3,-2)​

Answer»

Answer : We know that area of a triangle with vertices A(x'1,y'1),B(x'2,y'2) and C(x'3,y'3) is GIVEN by

ar (triangle ABC) = 1/2×[x'1(y'2 - y'3) + x'2(y'3 - y'1) + x'3(y'1 - y'2) ]

Hence , By the formula,

1/2×[2{1 - (-2)} + (-2){(-2) - 3} + 3(3 - 1)]

= 1/2×[ 6 + 10 + 6 ]

= 1/2× 22

= 11 sq units (Answer)

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