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.