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header-bg qa

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

Answers (1)

B)

\\ \begin{aligned} &\text { The area of a triangle with vertices }\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right) \text { is given by }\\ &\Delta=\frac{1}{2}\left|\begin{array}{lll} \mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3} & 1 \end{array}\right|\\ &\Delta=\frac{1}{2}\left|\begin{array}{ccc} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & \mathrm{k} & 1 \end{array}\right|=9\\ &\Delta=\left|\begin{array}{ccc} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & \mathrm{k} & 1 \end{array}\right|=18 \end{aligned}

Expand along Column 2

\vspace{\baselineskip} \Rightarrow -(k) \{ -3 -3 \} = 18\\ \vspace{\baselineskip} \Rightarrow -k (-6) = 18\\ \vspace{\baselineskip} \Rightarrow 6k = 18\\ \vspace{\baselineskip} \Rightarrow k = 3\\

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