(c) (5, 0) or (1, 4) Let the third vertex of the triangle be P(a, b). Since it lies on the line x + y = 5, a + b = 5 ...(i) Also, given area of triangle formed by the points (2, –1), (3, 2) and (a, b) = 4 units⇒ \(rac{1}{2}\)[2(2 – b) + 3(b + 1) + a(–1 – 2)] = ± 4 ⇒ [4 – 2b + 3b + 3 – 3a] = ± 8 ⇒ –3a + b = 1 ...(ii) or –3a + b = –15 ...(iii) Eqn (i) – Eqn (ii) ⇒ (a + b) – (–3a + b) = 5 – 1 ⇒ 4a = 4 ⇒ a = 1 ⇒ b = 4 Eqn (i) – Eqn (iii) ⇒ (a + b) – (–3a + b) = 5 + 15⇒ 4a = 20 ⇒ a = 5 ⇒ b = 0. ∴ The points are (1, 4) and (5, 0).