KCET · Maths · Straight Lines
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 \(6\)
- C \(3\)
- D \(9\)
Answer & Solution
Correct Answer
(C) \(3\)
Step-by-step Solution
Detailed explanation
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 \(C_2\)
\(9=\frac{1}{2}[-k(-3-3)]\)
\(\Rightarrow \quad 18=3 k+3 k=6 k\)
\(\therefore \quad k=\frac{18}{6}=3\)
\(\left(x_1, y_1\right),\left(x_2, y_2\right)\) 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 \(C_2\)
\(9=\frac{1}{2}[-k(-3-3)]\)
\(\Rightarrow \quad 18=3 k+3 k=6 k\)
\(\therefore \quad k=\frac{18}{6}=3\)
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