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Question
Find the area of ΔABC whose vertices are:
A(-5,7) , B (-4,-5) and C (4,5)
Solution
A(-5,7) , B (-4,-5) and C (4,5) are the vertices of . Δ ABC Then
`(x_1=-5, y_1=7) . (x_2= -4, y_2=-5) and (x_3=4,y_3=5)`
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 {(-5)(-5-5) +(-4)(5-7)+4(7-(5))}`
`=1/2 {(-5) (-10) -4(-2)+4(12)}`
`=1/2 {50+8+48}`
`=1/2 (106)`
= 53 sq . units
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