Question

If the area of a triangle $ABC$, with vertices $A(1,3),B(0$, 0) and $C(k,0)$ is $3sq$. units, then the value of $k$ is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (A)

Area of $\Delta ABC=3sq.$ units
$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}1& 3& 1\\ 0& 0& 1\\ k& 0& 1\end{array}\right|=\pm 3$
$\Rightarrow \left|\begin{array}{ccc}1& 3& 1\\ 0& 0& 1\\ k& 0& 1\end{array}\right|=\pm 6$
$\Rightarrow 1(0-0)-3(0-k)+1(0-0)=\pm 6$
$\Rightarrow 3k=\pm 6$
$\Rightarrow k=\pm 2$