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Question
Find the area of ΔABC whose vertices are:
A( 3,8) , B(-4,2) and C( 5, -1)
Solution
A( 3,8) , B(-4,2) and C( 5, -1) are the vertices of ΔABC . Then,
`(x_1 =3, y_1=8),(x_2=-4,y_2=2) and (x_3=5,y_3=-1)`
Area of triangle ABC
`=1/2 {x_1(y_2-y_3) + x_2(y_3-y_1) +x_3(y_1-y_2)}`
`=1/2 {3(2-(-1))+(-4)(-1-8)+5(8-2)}`
`=1/2 {3(2+1)-4(-9)+5(6)}`
`=1/2 {9+36+30}`
`=1/2(75)`
=37.5 sq . units
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