Question

The distance between the orthocentre and circumcentre of the triangle whose vertices are at $\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right),\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$ and (1, 0),
is __________.

Answer 1

In ∆ABC
$$\begin{gathered} A=\left(\frac{-1}{2},\frac{\sqrt{3}}{2}\right)B=\left(\frac{-1}{2},\frac{-\sqrt{3}}{2}\right)C=\left(1,0\right) \\ AB=\sqrt{{\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{-1}{2}+\frac{1}{2}\right)}^2} \\ \mathrm{i}.\mathrm{e}AB=\sqrt{{\left(\sqrt{3}\right)}^2}=\sqrt{3} \\ BC=\sqrt{{\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2} \\ \mathrm{i}.\mathrm{e}BC=\sqrt{\frac{3}{4}+\frac{9}{4}}=\sqrt{\frac{12}{4}}=\sqrt{3}\mathrm{and}AC=\sqrt{{\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(1+\frac{1}{2}\right)}^2}=\sqrt{3} \end{gathered}$$

Since AB = BC = AC
i.e ∆ABC is an equilateral triangle
⇒ Orthocentre and circumcentre of ∆ABC coincide
⇒ Distance between orthocentre and circumcentre of ∆ABC is 0.