Q94 / 180 · KCET 2023 · Mathematics

The area of a triangle with vertices $(-3,0)$, $(3,0)$ and $(0, k)$ is 9 sq units, find the value of $k$ is

  • A
    $-9$
  • B
    $6$
  • Correct answer
    C
    $3$
  • D
    $9$

Solution

We know that areas of a triangle with vertices $\left(x_1, y_1\right),\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ is given by $\begin{aligned} & \Delta=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right| \\ & \Delta=\frac{1}{2}\left|\begin{array}{ccc}-3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1\end{array}\right|\end{aligned}$ Expanding along $C_2$ $9=\frac{1}{2}[-k(-3-3)]$ $\Rightarrow \quad 18=3 k+3 k=6 k$ $\therefore \quad k=\frac{18}{6}=3$

Hence, the correct answer is option: 'C'.

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