Question

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

Answer 1

Area of the triangle formed by the vertices $\left(x_1,y_1\right),\left(x_2,y_2\right)\mathrm{and}\left(x_3,y_3\right)\mathrm{is}$ $\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$.

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

Therefore,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\left|x\left(4-5\right)+4\left(5-3\right)+3\left(3-4\right)\right| \\ \Rightarrow 4=\frac{1}{2}\left|x\left(-1\right)+\left(4\right)\left(2\right)+\left(3\right)\left(-1\right)\right| \\ \Rightarrow 4=\frac{1}{2}\left|-x+8-3\right| \\ \Rightarrow 4=\frac{1}{2}\left|-x+5\right| \\ \Rightarrow 8=\left|-x+5\right| \\ \Rightarrow -x+5=8\mathrm{or}-x+5=-8 \\ \Rightarrow -x=3\mathrm{or}-x=-13 \\ \Rightarrow x=-3\mathrm{or}x=13 \end{gathered}$$

Hence, ​the value of x is 13 and −3.