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Question
Find the value of k so that the area of the triangle with vertices A (k+1, 1), B(4, -3) and C(7, -k) is 6 square units
Solution
`"Let" A(x_1,y_1) = A(k+1,1) , B(x_2,y_2)= B (4,-3) and C(x_3,y_3) = C(7,-k) now`
`"Area "(Δ ABC) = 1/2 [x_1 (y_2-y_3) + x_2 (y_3-y_1) +x_3(y_1-y_2)}`
`⇒ 6=1/2 [(k+1) (-3+k)+4(-k-1) +7(1+3)]`
`⇒6=1/2[k^2 -2k-3-4k-4+28]`
`⇒ k^2-6k+9=0`
`⇒(k-3)^2 = 0⇒k=3`
Hence , k=3
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