If the area of the triangle with vertices (x, 3), (4, 4) and (3, 5) is 4 square units, find x.

Area of the triangle formed by the vertices $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is $\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$.

Now, the given vertices are (x, 3), (4, 4) and  (3, 5)
and the given area is 4 square units.

Therefore,

Area of triangle $=\frac{1}{2}|x(4-5)+4(5-3)+3(3-4)|$

$\Rightarrow 4=\frac{1}{2}|x(-1)+(4)(2)+(3)(-1)|$

$\Rightarrow 4=\frac{1}{2}|-x+8-3|$

$\Rightarrow 4=\frac{1}{2}|-x+5|$

$\Rightarrow 8=|-x+5|$

$\Rightarrow-x+5=8$ or $-x+5=-8$

$\Rightarrow-x=3$ or $-x=-13$

$\Rightarrow x=-3$ or $x=13$

Hence, ​the value of x is 13 and −3.