Question

An equilateral triangle $ABC$ has one vertex at $(2,0)$ and ortho center at $(1,\frac{1}{\sqrt{3}})$. Let $P=(\frac{1}{2},\frac{1}{4})$ and distances of $P$ from the sides $BC,AC,$ and $AB$ are $x,y$ and $z$ respectively, then $x+y+z$ is equal to

• A. $\frac{1}{\sqrt{3}}$
• B. $\sqrt{3}$
• C. 2
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is B

$\sqrt{3}$

In an equilateral triangle, the sum of three perpendicular lengths from any point inside the triangle is equal to the length of an altitude.

So, $x+y+z=AM$

Given, orthocentre, $H\equiv (1,\frac{1}{\sqrt{3}})$

$AH=\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt{3}}$

Since in an equilateral triangle, orthocentre, circumcentre and centroid coincide, so, $AM=\frac{3}{2}AH$

$\Rightarrow AM=\frac{3}{2}\times \frac{2}{\sqrt{3}}=\sqrt{3}$

Hence, $x+y+z=\sqrt{3}$