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Question
Find the area of ΔABC whose vertices are:
A(10,-6) , B (2,5) and C(-1,-3)
Solution
A(10,-6) , B (2,5) and C(-1,-3) are the vertex of ΔABC. Then,
`(x_1=10,y_1=-6),(x_2=2,y_2=5) and (x_3=-1,y_3=3)`
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{10(5-3)+2(3-(-6))+(-1)(-6-5)}`
`=1/2{10(2)+2(9)-1(-11)}`
`=1/2{20+18+11}`
`=1/2(49)`
=24.5 sq. units
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