Question

$A$ and $B$ are two fixed points whose co-ordinates respectively are $(3,2)$ and $(5,1)$. $ABP$ is an equilateral triangle on $AB$ situated on the side opposite to that of origin. Find the co-ordinates of $P$ and those of the orthocenter of triangle $ABP$.

Answer 1

Equation of $AB$ is $x=2y=7$ and its length $a=\sqrt{5}$, and mid-point of $AB$ is the point $L(4,3/2)$. If $P$ be the vertex of the equilateral triangle then its perpendicular distance $P$ from $AB$ if a $\sin {60}^o$
or $p=\sqrt{5}.\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{2}\sqrt{15}$
Also distance of orthocenter $H$ from $AB$ is $\frac{1}{3}p$.
Now both $H$ and $P$ lie on a line perpendicular to $AB$ whose slope will be $2$ and passing through $L(4,3/2)$,
$\therefore \tan \theta =2$ or $\sin \theta =2/\sqrt{5}$
and $\cos \theta =1/\sqrt{5}$
Hence $H$ and $P$ lie on
$\frac{x-4}{\cos \theta }=\frac{y-3/2}{\sin \theta }=p$ for $P$
and $=\frac{1}{3}p$ for $H$.
$x=4+p\cos \theta ,y=3/2+p\sin \theta$ for $P$
$x=4+\left(p/3\right)\cos \theta ,y=3/2+\left(p/3\right)\sin \theta$ for $H$
Putting the value of $p,\cos \theta$ and $\sin \theta$ in the above, we get
Point $p(4+\sqrt{3}/2,3/2+\sqrt{3})$
Point $H[5+\sqrt{3}/6,3/2+\left(1/2\right)\sqrt{3}]$