Question

The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be
(a) 9
(b) 3
(c) –9
(d) 6

Answer 1

Given: Area of a triangle with vertices (–3, 0), (3, 0) and (0, k) = 9 sq. units

Area of the triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is $\left|\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\right|$.

According to the question,
$$\begin{gathered} \left|\frac{1}{2}\left|\begin{array}{ccc}-3& 0& 1\\ 3& 0& 1\\ 0& k& 1\end{array}\right|\right|=9 \\ \Rightarrow \frac{1}{2}\left|\begin{array}{ccc}-3& 0& 1\\ 3& 0& 1\\ 0& k& 1\end{array}\right|=\pm 9 \\ \Rightarrow \left|\begin{array}{ccc}-3& 0& 1\\ 3& 0& 1\\ 0& k& 1\end{array}\right|=\pm 18 \\ \Rightarrow \left\{-3\left(0-k\right)-0\left(3\right)+1\left(3k\right)\right\}=\pm 18 \\ \Rightarrow \left\{-3\left(-k\right)+1\left(3k\right)\right\}=\pm 18 \\ \Rightarrow \left\{3k+3k\right\}=\pm 18 \\ \Rightarrow 6k=\pm 18 \\ \Rightarrow k=\pm 3 \end{gathered}$$

Hence, the correct option is (b).