KCET · Maths · Determinants
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
\[
\left|\begin{array}{ccc}
x_{1} & y_{1} & 4 \\
x_{2} & y_{2} & 4 \\
x_{3} & y_{3} & 4
\end{array}\right|^{2} \text { is }
\]
- A \( 32 \mathrm{k}^{2} \)
- B \( 16 \mathrm{k}^{2} \)
- C \( 64 \mathrm{k}^{2} \)
- D \( 48 \mathrm{k}^{2} \)
Answer & Solution
Correct Answer
(C) \( 64 \mathrm{k}^{2} \)
Step-by-step Solution
Detailed explanation
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
\[
\begin{array}{l}
\left|\begin{array}{lll}
x_{1} & y_{1} & 4 \\
x_{2} & y_{2} & 4 \\
x_{3} & y_{3} & 4
\end{array}\right|^{2}=16\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|^{2} \\
=16\left(4 k^{2}\right)=64 k^{2}
\end{array}
\]
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