Question

The area of a triangle is 5 sq. units. Two of its vertices are (2, 1) and (3, −2). The third vertex is (x, y) where y = x + 3. Find the coordinates of the third vertex.

Answer 1

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

Area of a triangle = 5 sq. units

– 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

∴ The coordinates of the third vertex is \((\frac{7}{2}, \frac{13}{2})\)