Question

What is the area of a triangle with vertices at (0, – 2), (8, – 2), and (9, 1)?

Answer 1

ANSWER:  

The area of the triangle with vertices at (0, – 2), (8, – 2), and (9, 1) is 12 square units.

SOLUTION:

Given, vertices of the triangle are A(0, -2), B(8, -2) and C(9, 1).

We have to find the area of the given triangle.

We know that,

$\text { Area of triangle }=\frac{1}{2}\left(x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right)$

$\text { Where, }\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right) \text { are vertices of the triangle. }$

Here in our problem, $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(0,-2),\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(8,-2) \text { and }\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right)=(9,1)$

Now, substitute the above values in the formula.

$\text { Area of triangle }=\frac{1}{2}\left(x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right)$

$=\frac{1}{2}(0(-2-1)+8(1-(-2))+9(-2-(-2))$

$=\frac{1}{2}(0+8(1+2)+9(2-2))$

$\frac{1}{2}(0 + 0 +24) = \frac{24}{12} = 2$

Hence, the area of the triangle is 12 sq units.