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. 

Answer 1

`"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