if the vertices of a triangle are 1 k 4 -3 -9 7

Question

If the vertices of a triangle are (1, k), (4, -3), (-9, 7) and its area is 15 sq units, find the value(s) of k..                                                                                                                                                      [Ans: -3 ,21/13 ]

Ans:    A(1, k)             B(4, -3)            C(-9, 7)

Area of Δ ABC =1/2 [x₁(y₂ -y₁ )+x₂ (y₃ -y₁ ) + x₁ (y₁ -y₂ )]

= 1/2 [1(-3-7)+4(7-k)+(-9)(k+3)] = 15

-10 + 28 - 4k - 9k - 27 = 30

- 9 - 13k = 30            ⇒        k = -3

| -9 - 13k | = 30

9 + 13k = 30

k = 21/13

k = -3, 21/13