Question

$A(1,3)$ and $B(0,0)$ are two points, using determinants and find $k$ if $D(k,0)$ is a point such that area of $△ABD$ is $3$ sq units.

• A. $k=\pm 3$
• B. $k=\pm 2$
• C. $k=3$
• D. $k=2$

Answer 1

The correct option is B $k=\pm 2$

Area of triangle$△\left(ABD\right)=\frac{1}{2}\left|\begin{array}{ccc}1& 3& 1\\ 0& 0& 1\\ k& 0& 1\end{array}\right|3=\frac{1}{2}\left[1\right(0-0)-3(0-k)+1(0-0\left)\right]3\times 2=\left|3k\right|k=\pm 2$