Question

The distance between the orthocentre and circumcentre of a triangle whose vertices are $P(3,0),Q(0,0)$ and $R(\frac{3}{2},\frac{-3\sqrt{3}}{2})$ is

• A. $0$
• B. $\sqrt{2}$
• C. $\sqrt{3}$
• D. $3\sqrt{3}$

Answer 1

The correct option is A $0$

$P(3,0),Q(0,0)$ and $R(\frac{3}{2},\frac{-3\sqrt{3}}{2})$
$\therefore PQ=\sqrt{(3-0{)}^2+(0-0{)}^2}=3$
$QR=\sqrt{{(\frac{3}{2}-0)}^2+{(\frac{-3\sqrt{3}}{2}-0)}^2}=3$
$PR=\sqrt{{(3-\frac{3}{2})}^2+{(0+\frac{3\sqrt{3}}{2})}^2}=3$

Hence, $PQ=QR=PR$
So, the triangle is equilateral.
Now, since in an equilateral triangle orthocentre and circumcentre coincides, therefore distance between them is zero.