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Determinants and Matrices — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDeterminants and Matrices2 Marks
Question
If the area of triangle with vertices $\mathrm{P}(-3,0), \mathrm{Q}(3,0)$ and $\mathrm{R}(0, \mathrm{~K})$ is 9 square unit then find the value of $k$.
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Answer

Given $\left(x_1, y_1\right) \equiv(-3,0),\left(x_2, y_2\right)$ $\equiv(3,0)$ and $\left(x_3, y_3\right) \equiv(0, \mathrm{k})$ and area of $\Delta$ is 9 sq. unit.
We know that area of $\Delta=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|$ $\therefore \pm 9=\frac{1}{2}\left|\begin{array}{ccc}-3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1\end{array}\right|$ (Area is positive but the determinant can be of either sign)
$
\begin{aligned}
& \therefore \pm 9=\frac{1}{2}[-3(0-k)+1(3 k-0)] \\
& \therefore \pm 9=\frac{1}{2}[3 \times 3 \mathrm{k}] \therefore \pm 9=3 k \quad \therefore \mathrm{k}= \pm 3
\end{aligned}
$

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