Two vertices of a

Let (h, k) be the third vertex of the triangle.

It is given that the area of the triangle with vertices (h, k), (− 2, − 1) and (3, 2) is 4 square units.

⇒ 3h – 5k + 1 = ± 8

Taking positive sign, we get,

3h – 5k + 1 = 8

3h – 5k – 7 = 0 … (1)

Taking negative sign, we get,

3h – 5k + 9 = 0 … (2)

The vertex (h, k) lies on the line x + y = 5.

H + k – 5 = 0 … (3)

On solving (1) and (3), we find (4, 1) to be the coordinates of the third vertex.

Similarly, on solving (2) and (3), we find (2, 3) to be the coordinates of the third vertex.