The area of the triangle with vertices $(\mathrm{P},0),(4,0),(0,2)$ is 4 square units, then find the value of P.

Here, the vertices of the triangle are  $(\mathrm{P},0),(4,0),(0,2)$.

Given, Area of the triangle is 4 square units.

We know, area of $\mathrm{\Delta ABC}=\frac{1}{2}\left|{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right|$.

where $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right),\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ and $\left({\mathrm{x}}_3,{\mathrm{y}}_3\right)$ are the vertices of the triangle.

On substitution,

$\Rightarrow \frac{1}{2}\left|\mathrm{P}\left(0-2\right)+4\left(2-0\right)+0\left(0-0\right)\right|=4$

$$$\Rightarrow \frac{1}{2}\left|-2\mathrm{P}+8\right|=4$

$\Rightarrow -2\mathrm{P}+8=4 \times 2$

$\Rightarrow -2\mathrm{P}+8=8$

$\Rightarrow -2\mathrm{P} = 0$

$\Rightarrow \mathrm{P} = 0$

Hence, the value of P is 0.