Question

Find the value of k, if area of triangle is 4 sq unit and vertices are

$(k,0),(4,0)(0,2)$

$(-2,0),(0,4),(0,k)$

Answer 1

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

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, $2k+8=-8\Rightarrow 2k=0\Rightarrow k=0$
$\therefore k=0,8$