Question

Let $A(1,3),B(0,0)$ and $C(k,0)$ be vertices of a triangle $ABC$ such that area of $△ABC$ is $3$. Find the value of $k$.

• A. $\pm 2$
• B. $\pm 3$
• C. $\pm 4$
• D. $\pm 1$

Answer 1

The correct option is A $\pm 2$

$A(1,3);B(0,0);C(k,0)$

Area$:\frac{1}{2}\left|△\right|$

$△=\left|\begin{array}{ccc}1& 1& 1\\ 1& 0& k\\ 3& 0& 0\end{array}\right|$

$△=1\left(0\right)-1(-3k)+0$

$△=3k$

Area$=\frac{1}{2}\left|△\right|=3$

$\Rightarrow \frac{1}{2}\left|3k\right|=3$

$\Rightarrow k=\pm 2$

Option A