Find the value of k for which the area formed by the triangle with vertices A(4,4) B(3,k) C(3,-2) is 7 sq.units
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A x ( B y − C y ) + B x ( C y − A y ) + C x ( A y − B y)
4{k-(-2)} + 3{(-2)-4} + 3{4-k}=7
4(k+2) + 3(-6) + 3(4-k)=7
4k+8-18+12-3k=7
k+22=7
k=7-22
k=-15
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