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

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

Determinants

Solution:

Area of $\Delta ABC =3 sq .$ units
$\Rightarrow \frac{1}{2} \begin{vmatrix}1&3&1\\ 0&0&1\\ k&0&1\end{vmatrix}=\pm3 $
$\Rightarrow \begin{vmatrix}1&3&1\\ 0&0&1\\ k&0&1\end{vmatrix}=\pm6$
$\Rightarrow 1(0-0)-3(0-k)+1(0-0)=\pm 6$
$\Rightarrow \quad 3 k=\pm 6$
$ \Rightarrow k=\pm 2$