Question

The distance between the orthocentre and circumcentre of the triangle with vertices $(1,2)(2,1)$ and $(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2})$ is

• A. 0
• B. $\sqrt{2}$
• C. $3+\sqrt{3}$
• D. none of these

Answer 1

The correct option is A

0

Let $A(1,2),B(2,1)$ and $C(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2})$

be the given points

$\therefore AB=\sqrt{(2-1{)}^2+(1-2{)}^2}$

$BC=\sqrt{{(\frac{3+\sqrt{3}}{2}-2)}^2+{(\frac{3+\sqrt{3}}{2}-1)}^2}$

$=\sqrt{2}$

$AC=\sqrt{{(\frac{3+\sqrt{3}}{2}-1)}^2+{(\frac{3+\sqrt{3}}{2}-2)}^2}$

$=\sqrt{2}$

Thus, ABC is an equilateral triangle. We know that the orthocentre and the circumcentre of an equilateral triangle are same.

So, the distance between the orthocentre and the circumcentre of the triangle with vertices $(1,2),(2,1)$ and $(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2})$ is 0.