#### 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*.

#### Solution 1

`therefore `

`rArr `

`rArr`

`rArr`

⇒ Area of ∆ABC = 4 sq. units

Hence, Area of ∆ABC is independent of t.

#### Solution 2

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 (*x*_{1}, *y*_{1}), (*x*_{2}, *y*_{2}) and (*x*_{3}, *y*_{3}) 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*.

Is there an error in this question or solution?

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