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If the area of a triangle with vertices $(-3,0),(3,0)$ and $(0, k)$ is 9 sq unit, then what is the value of $k ?$
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Let the vertices of the $\Delta \mathrm{ABC}$ be $\mathrm{A}(-3,0), \mathrm{B}(3,0)$ and $\mathrm{C}(0, \mathrm{k})$
$\therefore \quad$ Area of $\Delta \mathrm{ABC}=\frac{1}{2}\left|\begin{array}{ccc}-3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & \mathrm{k} & 1\end{array}\right|$
Given, area is 9
$\Rightarrow \quad 9=\frac{1}{2}\{-3(-\mathrm{k})+1(3 \mathrm{k})\}$
$\Rightarrow \quad 18=3 \mathrm{k}+3 \mathrm{k}$
$\Rightarrow \quad \mathrm{k}=\frac{18}{6}=3$
$\therefore \quad$ Area of $\Delta \mathrm{ABC}=\frac{1}{2}\left|\begin{array}{ccc}-3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & \mathrm{k} & 1\end{array}\right|$
Given, area is 9
$\Rightarrow \quad 9=\frac{1}{2}\{-3(-\mathrm{k})+1(3 \mathrm{k})\}$
$\Rightarrow \quad 18=3 \mathrm{k}+3 \mathrm{k}$
$\Rightarrow \quad \mathrm{k}=\frac{18}{6}=3$
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