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 [x_{1}(y_{2} -y_{1} )+x_{2} (y_{3} -y_{1} ) + x_{1} (y_{1} -y_{2} )]
= 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