Question

Find the values of $k$, if the area of triangle is $35sq.units$ whose vertices are $(2,-6),(5,4)$ and $(k,4)$.

Answer 1

Area of triangle is $△=\frac{1}{2}\left|\begin{array}{ccc}2& -6& 1\\ 5& 4& 1\\ k& 4& 1\end{array}\right|=35\left(given\right)$$△=(20-4k)-(8+6k)+(8+30)=70\left[openingfrom{C}_3\right]$$50-10k=70$$k=-2$
taking $+ve$ sign
$\Rightarrow -10k+50=70$
$\Rightarrow -10k=70-50$
$\Rightarrow -10k=20$
$\Rightarrow k=-2$
taking $-ve$ sign
$\Rightarrow -10k+50=-70$
$\Rightarrow -10k=-70-50$
$\Rightarrow -10k=-120$
$\Rightarrow k=12$
$\Rightarrow k=-2,12$