Question

The vertices of a triangle are $3+4i,4+3i$ and $2\sqrt{6}+i,$ then distance between ortho-centre and circum-centre of the triangle is equal to

• A. $\sqrt{137-28\sqrt{6}}$
• B. $\sqrt{137+28\sqrt{6}}$
• C. $\frac{1}{2}\sqrt{137+28\sqrt{6}}$
• D. $\frac{1}{3}\sqrt{137+28\sqrt{6}}$

Answer 1

Answer: (B)

Let,
$z_1=3+4i$ , $z_2=4+3i$ and $z_3=2\sqrt{6}+i$ are vertices of a triangle.
Here, $\left|z_1-z_0\right|=\left|z_2-z_0\right|=\left|z_3-z_0\right|=5$ (where, $z_0$ is the circum-centre situated at the origin).
Let, $z_G$ be the centroid then $z_G=\frac{z_1+z_2+z_3}{3}=\frac{7+2\sqrt{6}+8i}{3}$
Now, we know that the centroid, circum-centre and ortho-centre are collinear and


$\frac{HG}{GO}=\frac{2}{1}$
(where, $H\to$ ortho-centre, $G\to$ centroid and $O\to$ circum-centre).
Then,
$HG+GO=HO$
$\Rightarrow 2GO+GO=HO$
$\Rightarrow HO=3GO$
$\Rightarrow HO=3\left|\frac{7+2\sqrt{6}+8i}{3}-0\right|$
$\Rightarrow HO=\left|7+2\sqrt{6}+8i\right|$
$\Rightarrow HO=\sqrt{49+24+64+28\sqrt{6}}$
$\Rightarrow HO=\sqrt{137+28\sqrt{6}}$ .