The coordinates of the vertices of ∆ABC are A(4, 1), B(-3, 2) and C(0, k). Given that the area of ABC is 12 unit2, find the value of k.
Let the given points be A(1, 1), B(3, K) and C(-1, 4).
Here, we have
x1 = 1, y1 = 1
x2 = 3, y2 = k
and x3 = -1, y3 = 4
Given three points, will be collinear if,
x1 (y2 - y3) + x2(y3 -y1) + x3 (y1 - y2) = 0
⇒ 1(k -4) + 3(4 - 1) + (-1)(1 - k) = 0
⇒ k - 4 + 3 × 3 + (-1 + k) = 0
⇒ k - 4 + 9 - 1 + k = 0
⇒ 2K + 4 = 0
⇒ 2K = -4
Hence the value of K = -2