Question

If area of the triangle whose vertices are $(-3,0),$ $(3,0)$ and $(0,k)$ is $9$ sq. units, then $k$ can be

[1 mark]

• A. $4$
• B. $6$
• C. $3$
• D. $1$

Answer 1

The correct option is C $3$

We know the area of triangle whose vertices are $(x_1,y_1),$ $(x_2,y_2)$ and $(x_3,y_3)$ is
$\Delta =\left|\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\right|$

$\Rightarrow 9=\left|\frac{1}{2}\left|\begin{array}{ccc}-3& 0& 1\\ 3& 0& 1\\ 0& k& 1\end{array}\right|\right|$

$\Rightarrow 18=|-k(-3-3)|$
$\Rightarrow 18=|6k|$
$\Rightarrow k=\pm 3$