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