Question

The vertices of triangle are $A(I,2),B(2,I)$ & $C(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2})$ then distance between orthocentre & circumcentre is?

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

Answer 1

Answer: (B)

$0$

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

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

Distance(CA)=$\sqrt{{\left(\frac{\sqrt{3}+1}{2}\right)}^2+{\left(\frac{\sqrt{3}-1}{2}\right)}^2}=\sqrt{2}$

Therefore, it is an equilateral triangle.

So, the orthocenter and circumcenter coincide.

So, distance is zero.

Hence, 0 is the correct answer.