Question

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

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

Answer 1

The correct option is D $(4+\frac{1}{6}\sqrt{3},\frac{3}{2}+\frac{1}{3}\sqrt{3})$

Let M be the midpoint of line AB. Then

$M$ $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ $⟹$ $M$ $(4,\frac{3}{2})$

Slope of line AB $=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x1}=\frac{1-2}{5-3}=\frac{-1}{2}$

As the product of two perpendicular line is always $1$.

Then, Slope of line MP $\times$ Slope of line AB $=1$

Therefore, Slope of line MP $=-\frac{1}{-1/2}=2$

Slope $=\tan \theta$

$\Rightarrow \frac{\sin \theta }{\cos \theta }=2$

$\frac{\sin \theta }{\cos \theta }=\frac{2}{1}=\frac{\frac{2}{\sqrt{2^2+1^2}}}{\frac{1}{\sqrt{2^2+1^2}}}=\frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}$

$⟹\sin \theta =\frac{2}{\sqrt{5}}$ and $\cos \theta =\frac{1}{\sqrt{5}}$ -------------------eq(i)

Given triangle is an equilateral triangle.

So all sides and angles are equal and each angle equals to ${60}^o$

Side of the given triangle $=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}=\sqrt{(5-3{)}^2+(1-2{)}^2}=\sqrt{5}$

Let PM be $h$

PB as $a=\sqrt{5}$

In $△PBM$,

$\angle PBM={60}^o$

$\angle PMB={90}^o$

Hypotenuse $=a$

So, $\sin$ ${60}^o=\frac{h}{a}$

As $\sin \theta =\frac{perpendicular}{hypotenuse}$

$⟹h=a\times \sin {60}^o=\sqrt{5}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{15}}{2}$

For point P:

$\frac{x-4}{\cos \theta }=\frac{y-3/2}{\sin \theta }=h$

Here, $(4,3/2)$ are coordinates of point $M$

$\frac{y-3/2}{\sin \theta }=h=\frac{\sqrt{15}}{2}$

$⟹y=\frac{3}{2}+\sqrt{3}$

$\frac{x-4}{\cos \theta }=h=\frac{\sqrt{15}}{2}$

$⟹x=4+\frac{\sqrt{3}}{2}$

$P(4+\frac{\sqrt{3}}{2},\frac{3}{2}+\sqrt{3})$

Orthocenter $(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

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

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

Hence, option $D$ is correct.