If $p,q,r$ be lengths of perpendiculars from vertices of triangle $ABC$ on any line, prove $a^2(p-q)(p-r)+b^2(q-r)(q-p)+c^2(r-p)(r-q)=4\Delta^2$

Question

Let : $$A:=(x_1,y_1),$$ $$B:=(x_2,y_2),$$ $$C:=(x_3,y_3)$$ be the vertices of the triangle $ABC$. Consider an arbitrary straight line in perpendicular form $x\cos \theta + y\sin \theta - t = 0$. Then the lengths of the perpendiculars from the vertices of the triangle are: $$p=x_1\cos \theta + y_1\sin \theta - t,$$ $$q=x_2\cos \theta + y_2\sin \theta - t,$$ $$r=x_3\cos \theta + y_3\sin \theta - t.$$ Moreover, the lengths of the sides of the triangle are: $$a^2=(x_3-x_2)^2+(y_3-y_2)^2,$$ $$b^2=(x_1-x_3)^2+(y_1-y_3)^2,$$ $$c^2=(x_2-x_1)^2+(y_2-y_1)^2.$$ Using the above values I tried to evaluate the LHS to prove the desired result, but that way is too tedious. Can anyone suggest me a better proof?

Answer 1

Hint. Let $$f(u,v,w):=(v-w)^2(p-q)(p-r)+(w-u)^2(q-r)(q-p)+(u-v)^2(r-p)(r-q)\,.$$ Observe that $$\begin{align}f(u,v,w)&=\big((v-w)p\big)^2+\big((w-u)q\big)^2+\big((u-v)r\big)^2\\&\phantom{aaaaa}+2\,\big((w-u)q\big)\,\big((u-v)r\big)+2\,\big((u-v)r\big)\,\big((v-w)p\big)+2\,\big((v-w)p\big)\,\big((w-u)q\big)\,.\end{align}$$ Therefore, $$f(u,v,w)=\Big(\det\big(M(u,v,w)\big)\Big)^2\,,$$ where $$M(u,v,w):=\begin{bmatrix}1&p&u\\1&q&v\\1&r&w\end{bmatrix}\,.$$