The area of a triangle is 5. Two of its vertices are (2, 1) and (3, - 2). The third vertex lies on y = x + 3. Find the third vertex.

Given: vertices (2, 1) and (3, - 2).
y = x + 3

To find: The third vertex

Formula Used:

Area of the triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃)
= 1/2 |x₁(y₂ – y₃) + x₂(y₃ - y₁) + x₃(y₁ – y₂) |

Explanation:

Let ABC be a triangle with A (a, b),B(2,1) and C(3, - 2).
A lies on the line y = x + 3 means,
b = a + 3

b – a = 3 ...(1).

Area of ∆ABC = 5

Substituting the values of A, B and C in the formula, we, get,
5 = 1/2 | 3a + b – 7 |

Taking positive value for | 3a + b – 7 |,
3a + b = 17 …. (2)

Subtract (1) from (2) to get,

3a + b – (b – a) = 17 – 3

⇒ 3a + b – b + a = 14

⇒ 4a = 14

Put the value of a in (1) to get,

So,
Hence coordinates of the vertex A are

Taking negative value for | 3a + b – 7 |,
1/2 (3a + b−7) = − 5

⇒ 3a + b−7 = − 10

⇒ 3a + b = - 10 + 7

⇒ 3a + b = −3 … (3)

Subtract 1 from 3 to get,

3a + b – (b – a) = - 3 - 3

⇒ 3a + b – b + a = - 6

⇒ 4a = - 6

Put this value in 1 to get,

and the vertex A is

Hence the coordinates of the third vertex are or