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  4. If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of an equilateral triangle whose each side is equal to a, then | beginmatrixx1&y1&4 x2&y2&4 x3&y3&4 endmatrix|2 is equal to

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Q. If $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$ and $C\left(x_{3}, y_{3}\right)$ are the vertices of an equilateral triangle whose each side is equal to a,
then $\left|\begin{matrix}x_{1}&y_{1}&4\\ x_{2}&y_{2}&4\\ x_{3}&y_{3}&4\end{matrix}\right|^{2}$ is equal to

Determinants

Solution:

Let $\Delta$ be the area of triangle $ABC$. Then,

$\Delta=\frac{1}{2}\left|\begin{matrix}x_{1}&y_{1}&1\\ x_{2}&y_{2}&1\\ x_{3}&y_{3}&1\end{matrix}\right|\Rightarrow 2\Delta=\left|\begin{matrix}x_{1}&y_{1}&1\\ x_{2}&y_{2}&1\\ x_{3}&y_{3}&1\end{matrix}\right|$

$\Rightarrow 8 \Delta=4 \left|\begin{matrix}x_{1}&y_{1}&1\\ x_{2}&y_{2}&1\\ x_{3}&y_{3}&1\end{matrix}\right|=\left|\begin{matrix}x_{1}&y_{1}&4\\ x_{2}&y_{2}&4\\ x_{3}&y_{3}&4\end{matrix}\right|$

$\Rightarrow \quad64 \Delta^{2}=\left|\begin{matrix}x_{1}&y_{1}&4\\ x_{2}&y_{2}&4\\ x_{3}&y_{3}&4\end{matrix}\right|^{2}\quad\ldots\left(i\right)$

But, the area of an equilateral triangle with each side $a$ is $\frac{\sqrt{3}}{4} a^{2}$.

$\therefore \quad\Delta=\frac{\sqrt{3}}{4}a^{2} \Rightarrow 16 \Delta^{2}=3a^{4} \quad\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$, we get

$\left|\begin{matrix}x_{1}&y_{1}&4\\ x_{2}&y_{2}&4\\ x_{3}&y_{3}&4\end{matrix}\right|^{{2}}=12 a^{4}$