Question

The vertices of $△ABC$ are $(-2,1),(5,4)$ and $(2,-3)$ respectively. Find the area of the triangle and the length of the altitude through $A$.

Answer 1

We will use the formula to find the area of the triangle when all three coordinates are given,

$\begin{array}{cc}A& =\frac{1}{2}\left[y_1\left(x_2-x_3\right)+y_2\left(x_3-x_1\right)+y_3\left(x_1-x_2\right)\right]\\ & =\frac{1}{2}\left[1\left(5-2\right)+4\left(2+2\right)-3\left(-2-5\right)\right]\\ & =\frac{1}{2}\left[3+4\times 4-3\times -7\right]\\ & =\frac{1}{2}\left(3+16+21\right)\\ & =\frac{1}{2}\times 40\\ & =20\text{sq. units}\end{array}$

But as we know the area of a triangle is equal to half of the product of base and height. So, if we assume base as $BC$ then the height will be equal to the length of the altitude from the vertex $A.$

$\begin{array}{cc}BC& =\sqrt{{\left(2-5\right)}^2+{\left(-3-4\right)}^2}\\ & =\sqrt{{\left(-3\right)}^2+{\left(-7\right)}^2}\\ & =\sqrt{9+49}\\ & =\sqrt{58}\text{units}\end{array}$

Now,

$\begin{array}{cc}Area& =\frac{1}{2}\times BC\times AD\\ 20& =\frac{1}{2}\times \sqrt{58}\times AD\\ AD& =\frac{40}{\sqrt{58}}\text{units}\end{array}$