The two vertices of a triangle are (2, –1), (3, 2) and the third ver

1.	The two vertices of a triangle are (2, –1), (3, 2) and the third vertex lies on the line x + y = 5. The area of the triangle is 4 units. The third vertex is(a) (0, 5) or (1, 4) (b) (5, 0) or (4, 1) (c) (5, 0) or (1, 4) (d) (0, 5) or (4, 1)
Answer» (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 ⇒ \(\frac{1}{2}\)[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).

The two vertices of a triangle are (2, –1), (3, 2) and the third vertex lies on the line x + y = 5. The area of the triangle is 4 units. The third vertex is(a) (0, 5) or (1, 4) (b) (5, 0) or (4, 1) (c) (5, 0) or (1, 4) (d) (0, 5) or (4, 1)

Answer»

(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

⇒ \(\frac{1}{2}\)[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).

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