Question

Find area of the triangle formed by the points $\left(2,3\right),\left(6,3\right)$ and $\left(2,6\right)$ using Heron's formula

Answer 1

Solve for area of the required triangle

Let the vertices of triangle be,

$$\begin{gathered} A\left(x_1,y_1\right)=\left(2,3\right) \\ B\left(x_2,y_2\right)=\left(6,3\right) \\ C\left(x_3,y_3\right)=\left(2,6\right) \end{gathered}$$

Step 1: Solve for lengths of $AB,BC$ and $AC$

We know that,

Distance between $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$$\begin{gathered} \Rightarrow AB=c=\sqrt{{\left(6-2\right)}^2+{\left(3-3\right)}^2}=4units \\ \Rightarrow BC=a=\sqrt{{\left(2-6\right)}^2+{\left(6-3\right)}^2}=5units \\ \Rightarrow AC=b=\sqrt{{\left(2-2\right)}^2+{\left(6-3\right)}^2}=3units \end{gathered}$$

Step 2: Solve for area of triangle

Using Heron's formula

Area of triangle$\mathbf{=}\sqrt{\mathbf{s}\left(s-a\right)\left(s-b\right)\left(s-c\right)}$, where $\mathit{s}\mathbf{=}\frac{\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}}{\mathbf{2}}$

Here, $s=\frac{5+3+4}{2}=6units$

$\therefore$Area of triangle$=\sqrt{6\left(6-5\right)\left(6-3\right)\left(6-4\right)}=\sqrt{6\times 1\times 3\times 2}$

$\Rightarrow$Area of the given triangle $=6Sq.units$

Hence, area of the triangle with vertices $(2,3),(6,3)$ and $\mathrm{(}\mathrm{2}\mathrm{,}\mathrm{6}\mathrm{)}$ is $6sq.units$