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.
Given points are A(-5, 7), B(-4, 5) and C(4, 5).
Here, we have
x1 = -5, y1 = 7
x2 = -4, y2 -5
and x3 = 4, y3 = 5
Area of ∆ABC
Hence area of the ∆ABC = 53 sq. units.