If area of triangle is 4 sq units with vertices (-2, 0), (0, 4) and (0, k), then k is equal to

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

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}|-2(4-k)+1(0-0\left)\right|=8\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-2(4-k)+1(0-0)=\pm 8\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(-8+2k)=\pm 8\end{array}$

Taking positive sign,

$\begin{array}{r}2k-8=8\\ \Rightarrow 2k=16\Rightarrow k=8\end{array}$

Taking negative sign,

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\mathrm{k}-8=-8\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\mathrm{k}=0\Rightarrow \mathrm{k}=0\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{k}=0,8\end{array}$