Question

Let $A\equiv (3,2)$ and $B\equiv (5,1)$. $ABP$ is an equilateral triangle constructed on the side of $AB$ remote from the origin then the orthocentre of triangle $ABP$ is

• A. $(4+\frac{\sqrt{3}}{6},\frac{3}{2}+\frac{\sqrt{3}}{3})$
• B. $(4+\frac{1}{2}\sqrt{3},\frac{3}{2}+\sqrt{3})$
• C. $(4+\frac{1}{6}\sqrt{3},\frac{3}{2}-\frac{1}{3}\sqrt{3})$
• D. $(4+\frac{1}{6}\sqrt{3},\frac{3}{2}+\frac{1}{3}\sqrt{5})$

Answer 1

The correct option is A $(4+\frac{\sqrt{3}}{6},\frac{3}{2}+\frac{\sqrt{3}}{3})$

$A=(3,2)B=(s,1)$

$m_{AB}=-1/2$

$\because P_m$ is $h$ to $AB$

$\therefore m_{pm}=2$

$h=\frac{a\sqrt{3}}{2}$ now $a=\sqrt{5}=AB,n=\sqrt{15}/2$

So point $P\left(\frac{x-y}{\cos \theta }\right)=\frac{y-3/2}{\sin \theta }=n$

$P=:x=4+\sqrt{3}/2,y=3/2+\sqrt{3}$

So, orthocentre /centroid $\Rightarrow 4+\frac{\sqrt{3}}{6},3/2+\sqrt{3}/3$