ExamBro
ExamBro
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
\[
\left|\begin{array}{ccc}
x_{1} & y_{1} & 4 \\
x_{2} & y_{2} & 4 \\
x_{3} & y_{3} & 4
\end{array}\right|^{2} \text { is }
\]

  1. A \( 32 \mathrm{k}^{2} \)
  2. B \( 16 \mathrm{k}^{2} \)
  3. C \( 64 \mathrm{k}^{2} \)
  4. D \( 48 \mathrm{k}^{2} \)
Verified Solution

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
\[
\begin{array}{l}
\left|\begin{array}{lll}
x_{1} & y_{1} & 4 \\
x_{2} & y_{2} & 4 \\
x_{3} & y_{3} & 4
\end{array}\right|^{2}=16\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|^{2} \\
=16\left(4 k^{2}\right)=64 k^{2}
\end{array}
\]