Question

Find the value of k if area of triangle is $4$ square units and vertices are $(k,0),(4,0)$ and $(0,2)$

Answer 1

The area of the triangle with vertices $(k,0),(4,0)$ and $(0,2)$ is given by the relation,

$=\frac{1}{2}\left|\begin{array}{ccc}k& 0& 1\\ 4& 0& 1\\ 0& 2& 1\end{array}\right|$

$\Rightarrow \frac{1}{2}[-2k+8]=\pm 4$

$\Rightarrow -k+4=\pm 4$

Hence, $k=0,8$.