Q106 / 180 · KCET 2018 · Mathematics
If \( \left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right) \) and \( \left(x_{3}, y_{3}\right) \) are the vertices of a triangle whose area is ' \( k \) ' square units,
then
\[
\left|\begin{array}{ccc}
x_{1} & y_{1} & 4 \\
x_{2} & y_{2} & 4 \\
x_{3} & y_{3} & 4
\end{array}\right|^{2} \text { is }
\]
then
\[ \left|\begin{array}{ccc} x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4 \end{array}\right|^{2} \text { is } \]
- A\( 32 \mathrm{k}^{2} \)
- B\( 16 \mathrm{k}^{2} \)
- Correct answerC\( 64 \mathrm{k}^{2} \)
- D\( 48 \mathrm{k}^{2} \)
Solution
Vertices of a triangle are \( \left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right) \) and \( \left(x_{3}, y_{3}\right) . \) So, area of triangle is given by,
\( \frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|=k \)
We have
\[ \begin{array}{l} \left|\begin{array}{lll} x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4 \end{array}\right|^{2}=16\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|^{2} \\ =16\left(4 k^{2}\right)=64 k^{2} \end{array} \]
\( \frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|=k \)
We have
\[ \begin{array}{l} \left|\begin{array}{lll} x_{1} & y_{1} & 4 \\ x_{2} & y_{2} & 4 \\ x_{3} & y_{3} & 4 \end{array}\right|^{2}=16\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|^{2} \\ =16\left(4 k^{2}\right)=64 k^{2} \end{array} \]
Hence, the correct answer is option: 'C'.