Question

Prove that the area of a triangle with vertices (t, t −2), (t + 2, t + 2) and (t + 3, t) is independent of t.

Answer 1

Let A(t, t − 2), B(t + 2, t + 2) and C(t + 2, t) be the vertices of the given triangle.

We know that the area of the triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

`1/2|x_1(y_2-y_3)+x^2(y_3-y_1)+x_3(y_1-y_2)||`

∴ Area of ∆ABC =  `|1/2[x_1(y_2-y_3)+x^2(y_3-y_1)+x_3(y_1-y_2)||`

`=|1/2[t(t+2−t)+(t+2)(t−t+2)+(t+3)(t−2−t−2)]|`

`|1/2(2t+2t+4−4t−12)|`

`|-4|`

=4 square units

Hence, the area of the triangle with given vertices is independent of t.

Answer 2

`therefore `

`rArr `

`rArr`

`rArr`

⇒ Area of ∆ABC = 4 sq. units

Hence, Area of ∆ABC is independent of t.