Question

If a vertex of an equilateral triangle is the origin and the side opposite to it has the equation $x+y=1$ then the orthocentre of the triangle is

• A. $\left(\frac{1}{3},\frac{1}{3}\right)$
• B. $\left(\frac{\sqrt{2}}{3},\frac{\sqrt{2}}{3}\right)$
• C. $\left(\frac{2}{3},\frac{2}{3}\right)$
• D. None of these

Answer 1

Answer: (B)

Clearly orthocentre ' $H$ ' lies on the line $x-y=0$.
Now distance of $O(0,0)$ from the line $x+y-1=0$ is $\frac{1}{\sqrt{2}}$.
$\therefore OH=\frac{2}{3}\cdot \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{3}$ (since triangle is equilateral, centroid coincides with orthocentre)
$\therefore$ orthocentre $\equiv \left(\frac{\sqrt{2}}{3},\frac{\sqrt{2}}{3}\right)$