Join two points $\mathrm{P}(2, 2)$ and $\mathrm{Q}(4, 2)$ in a plane. Fix the point P and rotate the line PQ in anti-clockwise direction at an angle of $270°$. Find out the area formed by this figure.

Given points in a plane are $\mathrm{P}(2, 2)$ and $\mathrm{Q}(4, 2)$.
By fixing the point P, by rotating the line Q in an anti-clockwise direction at an angle  we get the following figure,
.
Let Q' be the new coordinates of Point Q then,
$\mathrm{Q}\text{'}(2,0)$ .
From the graph, it is clear that we have to find out the area of a triangle.
For this,
Since, $\angle \mathrm{Q}\text{'}\mathrm{PQ}=90°$   {From graph}.
$\therefore ∆\mathrm{QPQ}\text{'}$ is a right-angled triangle.
Also,
$\mathrm{PQ}\text{'}=\mathrm{PQ}=2 \mathrm{units}$    {From graph}.
We know that,
$\mathbf{Area}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Triangle}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\times }\mathbf{Base}\mathbf{\times }\mathbf{Height}$.
Then,
$\mathrm{ar}\left(∆\mathrm{QPQ}\text{'}\right)=\frac{1}{2}\times \mathrm{PQ}\text{'}\times \mathrm{PQ}$ 

$\Rightarrow \mathrm{ar}\left(∆\mathrm{QPQ}\text{'}\right)=\frac{1}{2}\times 2\times 2$ 

$\mathbf{\Rightarrow }\mathbf{ar}\left(∆\mathrm{QPQ}\text{'}\right)\mathbf{=}\mathbf{2}\mathbf{ }{\mathbf{units}}^{\mathbf{2}}$ 

Therefore, area of the figure is $\mathbf{2}\mathbf{ }{\mathbf{units}}^{\mathbf{2}}$.