Question

Find area of the triangle whose vertices are $\left(2,3\right),\left(-1,0\right)$ and $\left(2,-4\right)$

Answer 1

Solve for area of triangle

Let the vertices of triangle be,

$$\begin{gathered} A\left(x_1,y_1\right)=\left(2,3\right) \\ B\left(x_2,y_2\right)=\left(-1,0\right) \\ C\left(x_3,y_3\right)=\left(2,-4\right) \end{gathered}$$

We know that,

Area of the triangle$\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

$\therefore$Area of the required triangle $=\frac{1}{2}\left|2\left(0+4\right)-1\left(-4-3\right)+2\left(3-0\right)\right|$

$=\frac{1}{2}\left|8+7+6\right|=10.5sq.units$

Hence, area of the triangle with vertices $(2,3),(-1,0)$ and $\mathrm{(}\mathrm{2}\mathrm{,}\mathrm{-}\mathrm{4}\mathrm{)}$ is $10.5sq.units$