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 ${\left|\begin{array}{ccc}{x}_1& y_1& 2\\ x_2& y_2& 2\\ x_3& y_3& 2\end{array}\right|}^2$ 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,

$\Delta =\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

$\Rightarrow 2\Delta =\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\Rightarrow 4\Delta =2\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

$\Rightarrow 4\Delta =\left|\begin{array}{ccc}{x}_1& y_1& 2\\ x_2& y_2& 2\\ x_3& y_3& 2\end{array}\right|\Rightarrow 16{\Delta }^2={\left|\begin{array}{ccc}{x}_1& y_1& 2\\ x_2& y_2& 2\\ x_3& y_3& 2\end{array}\right|}^2\dots \left(i\right)$

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

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

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

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

$\Rightarrow 16{\Delta }^2=\left(p+q+r\right)\left(q+r-p\right)\left(r+p-q\right)\left(p+q-r\right)\dots \left(ii\right)$

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

${\left|\begin{array}{ccc}{x}_1& y_1& 2\\ x_2& y_2& 2\\ x_3& y_3& 2\end{array}\right|}^2=\left(p+q+r\right)\left(q+r-p\right)\left(r+p-q\right)\left(p+q-r\right)$