Answer 1

Let the vertices A(2, 1), B(3, –2) and C(x, y)

Area of a triangle = 5 sq. units

`1/2[x_1y_2 + x_2y_3 + x_3y_1 - (x_2y_1 + x_3y_2 + x_1y_3)]` = 5

`1/2[- 4 + 3y + x - (3 - 2x + 2y)]` = 5

– 4 + 3y + x – 3 + 2x – 2y = 10

3x + y – 7 = 10

3x + y = 17  ...(1)

Given y = x + 3

Substitute the value ofy = x + 3 in (1)

3x + x + 3 = 17

4x = 17 – 3

4x = 14

x = `14/4 = 7/2`

Substitute the value of x in y = x + 3

y = `7/2 + 3`

⇒ y = `(7 + 6)/2`

= `13/2`

∴ The coordinates of the third vertex is `(7/2, 13/2)`