Question

Find the value of k, if area of triangle is 4 sq unit and vertices are
$(-2,0),(0,4),(0,k)$

Answer 1

Given, $(\frac{1}{2}\left|\begin{array}{ccc}-2& 0& 1\\ 0& 4& 1\\ 0& k& 1\end{array}\right|=4\Rightarrow |-2(4-k)+1(0-0)|=8$
$\Rightarrow -2(4-k)+1(0-0)=\pm 8\Rightarrow [-8+2k]=\pm 8$
On taking positive sign, $2k-8=8\Rightarrow 2k=16\Rightarrow k=8$
On taking negative sign, $2k+8=-8\Rightarrow 2k=0\Rightarrow k=0$
$\therefore k=0,8$