Question

Distance between the circumcentre and orthocentre of $△ABC$ is

• A. $5$
• B. $\sqrt{29}$
• C. $7$
• D. $\sqrt{37}$

Answer 1

Answer: (C)

$7$

$\because A\equiv (1,2),B\equiv (2,3)$ and $C\equiv (4,3)$
$\therefore$ Centroid $G\equiv (\frac{1+2+4}{3},\frac{2+3+3}{2})=(\frac{7}{3},\frac{8}{3})$
Let $O^{\prime }(\alpha ,\beta )$ be the orthocentre
As $O^{\prime }D\perp$ to $BC$ we have
$\therefore$ Slope of $O^{\prime }A\times$Slope of $BC=-1$
$\Rightarrow \frac{\beta -2}{\alpha -1}\times \frac{0}{2}=-1$
or $\alpha -1=0$
$\therefore \alpha =1$
As $O^{\prime }E\perp$ to $AB$ we have
Also, slope of $O^{\prime }C\times$slope of $AB=-1$
$\Rightarrow \frac{\beta -3}{\alpha -4}\times \frac{3-2}{2-1}=-1$
Substituting for $\alpha =1$ from above we get
$\Rightarrow \beta -3=-(\alpha -4)=-1(1-4)=3$
$\therefore \beta =3+3=6$
$\therefore$ Orthocentre $O^{\prime }\equiv (1,6)$
$\because$ Centroid $G$ divides orthocentre $O^{\prime }$ and circumcentre $C^{\prime }$ in the ratio $2:1$ internally.
If $C^{\prime }\equiv (x_1,y_1)$ then
$O^{\prime }(1,6):G(\frac{7}{3},\frac{8}{3})=2:1$
$G(\frac{7}{3},\frac{8}{3}):C^{\prime }(x_1,y_1)=1:1$
$\therefore G$ divides the line $O^{\prime }{C}^{\prime }$ in the ratio $2:1$ which is given by the section formula: $(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$
$\therefore \frac{7}{3}=\frac{2.x_1+1.1}{2+1}$
$\Rightarrow x_1=3$ and $\frac{8}{3}=\frac{2.y_1+1.6}{2+1}$
On simplification, we get
$2y_1+6=8$ or $y_1=1$
$\therefore C^{\prime }\equiv (3,1)$
Then , $O^{\prime }{C}^{\prime }=|1-3|+|6-1|=2+5=7$units