For what values of k are the points A(8, 1), B( 3, –2k) and C(k, –5) collinear.
To show that the points are collinear, we show that the area of triangle is equilateral = 0
Δ = 0
Δ = 1/2{x1(y2−y3) + x2(y3−y1) + x3(y1−y2)}
⇒ Δ = 1/2{8(–2k + 5) + 3 (–5–1) + k (1 + 2k)} = 0
⇒ –16k + 40–18 + k + 2k2 = 0
⇒ 2k2 + 15k + 22 = 0
⇒ 2k2–11k–14k + 22 = 0
⇒ K(2k–11)–2(2k–11) = 0
k = 2 or k = 
18