Question

A triangle has its three sides equal to p, q and r. If the coordinates of its vertices are $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$ and $C\left(x_{3}, y_{3}\right)$, then $\left|\begin{matrix}x_{1}&y_{1}&2\\ x_{2}&y_{2}&2\\ x_{3}&y_{3}&2\end{matrix}\right|^{2}$ is equal to

• A. $\left(p+q+r\right)$
• B. $\left(p+q-r\right)^{2}$
• C. $\left(p+q+r\right) \left(q+r-p\right)\left(r+p-q\right)\left(p+q-r\right)$
• D. $\left(p-q+r\right) \left(q+r-p\right)\left(p+q-r\right)$

Answer 1

Answer: (C)

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

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

$\Rightarrow \quad2\Delta=\left|\begin{matrix}x_{1}&y_{1}&1\\ x_{2}&y_{2}&1\\ x_{3}&y_{3}&1\end{matrix}\right| \Rightarrow 4\Delta=2\left|\begin{matrix}x_{1}&y_{1}&1\\ x_{2}&y_{2}&1\\ x_{3}&y_{3}&1\end{matrix}\right|$

$\Rightarrow \quad4\Delta=\left|\begin{matrix}x_{1}&y_{1}&2\\ x_{2}&y_{2}&2\\ x_{3}&y_{3}&2\end{matrix}\right|\Rightarrow 16\Delta^{2}=\left|\begin{matrix}x_{1}&y_{1}&2\\ x_{2}&y_{2}&2\\ x_{3}&y_{3}&2\end{matrix}\right|^{2} \quad\ldots\left(i\right)$

We also know that the area of triangle ABC is given by

$\Delta=\sqrt{s\left(s-p\right)\left(s-q\right)\left(s-r\right)}, where s =\frac{1}{2}\left(p+q+r\right)$

But, $s=\frac{1}{2} \left(p+q+r\right)\Rightarrow s-p=\frac{1}{2}\left(p+q+r\right)-p=\frac{1}{2}\left(q+r-p\right)$

Similarly, $s-q=\frac{1}{2}\left(r+p-q\right)and s - r =\frac{1}{2}\left(p+q-r\right)$

$\therefore \Delta^{2}=\frac{1}{2}\left(p+q+r\right)\cdot\frac{1}{2}\left(q+r-p\right)\cdot\frac{1}{2} \left(r+p-q\right)\cdot\frac{1}{2}\left(p+q-r\right)$

$\Rightarrow \quad16\Delta^{2}=\left(p+q+r\right)\left(q+r-p\right)\left(r+p-q\right)\left(p+q-r\right)\quad\ldots\left(ii\right)$

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

$\left|\begin{matrix}x_{1}&y_{1}&2\\ x_{2}&y_{2}&2\\ x_{3}&y_{3}&2\end{matrix}\right|^{2}=\left(p + q + r\right)\left(q+r-p\right)\left(r+p-q\right)\left(p+q-r\right)$