Question

$A$ and $B$ are two fixed points whose coordinates are $(3,2)$ and $(5,1)$ respectively; $ABP$ is an equilateral triangle on the side of $AB$ remote from the origin. Find the coordinates of $P$ and the ortho-centre of the triangle $ABP$.

Answer 1

Draw the perpendicular $PM$

Then coordinates of $M$ are $(\frac{3+5}{2},\frac{2+1}{2})$

$AB=a=\sqrt{{(5-3)}^2+{(1-2)}^2}=\sqrt{5}$

In $△PMB$

${PM}^2+{MB}^2={BP}^2{h}^2+{\left(\frac{a}{2}\right)}^2=a^2\Rightarrow h=\frac{\sqrt{3}a}{2}h=\frac{\sqrt{3}\times \sqrt{5}}{2}=\frac{\sqrt{15}}{2}$

$\Rightarrow M(4,\frac{3}{2})$

Slope of $AB=\frac{1-2}{5-3}=-\frac{1}{2}$

$\Rightarrow$ Slope of $PM=2$

$\tan \theta =2$

$\sec \theta =\sqrt{1+{\tan }^2\theta }=\sqrt{1+4}=\sqrt{5}$

$\Rightarrow \cos \theta =\frac{1}{\sqrt{5}}$

$\tan \theta =\frac{\sin \theta }{\cos \theta }$

$\Rightarrow \sin \theta =\frac{2}{\sqrt{5}}$

$\frac{x-4}{\cos \theta }=\frac{y-\frac{3}{2}}{\sin \theta }=h\Rightarrow x=h\cos \theta +4=\frac{\sqrt{15}}{2}\times \frac{1}{\sqrt{5}}+4=\frac{\sqrt{3}}{2}+4y=h\sin \theta +\frac{3}{2}=\frac{\sqrt{15}}{2}\times \frac{2}{\sqrt{5}}+\frac{3}{2}=\sqrt{3}+\frac{3}{2}$

So, the coordinates of $P$ are $(\frac{\sqrt{3}}{2}+4,\sqrt{3}+\frac{3}{2})$

As the triangle is equilateral so the ortho-centre of triangle will be same as centroid. So the ortho-centre of the $△APB$ is

$(\frac{(3+5+\frac{\sqrt{3}}{2}+4)}{3},\frac{(2+1+\sqrt{3}+\frac{3}{2})}{3})(4+\frac{\sqrt{3}}{6},\frac{3}{2}+\frac{\sqrt{3}}{3})$