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 $z_{G}=\frac{z_{1} + z_{2} + z_{3}}{3}=\frac{7 + 2 \sqrt{6} + 8 i}{3}$
Now, we know that the centroid, circum-centre and ortho-centre are collinear and


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