If the points A(2, 3), B(4, k) and C(6, -3) are collinear, find the value of k.
If the three points are collinear then the area of the triangle formed by them will be zero.
Area of a Δ ABC whose vertices are A(x1,y1), B(x2,y2) and C(x3,y3) is given by-
∴ Area of given Δ ABC = 0
⇒ √(2(k-(-3)) + 4(-3-3) + 6(3-k) ) = 0
squaring both sides, we get-
2(k + 3) + 4(-6) + 6(3-k) = 0
⇒ 2k + 6-24 + 18-6k = 0
⇒ -4k + 24-24 = 0
∴ k = 0
Thus, the value of k is zero.