Question

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

Answer 1

$OD=|-\frac{1}{\sqrt{2}}|=\frac{1}{\sqrt{2}}$

$\tan 60=\frac{OD}{DB}$

$\Rightarrow DB=\frac{OD}{\sqrt{3}}=\frac{1}{\sqrt{6}}$

$\Rightarrow AB=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}$, Slope $x+y=1$

$\Rightarrow =-1$

Slope of OD$=1$

$\Rightarrow \angle DOQ={45}^o$

$\Rightarrow OQ=QD=QD=1$

$\sin 45=\frac{1}{DM}$

$\Rightarrow DM=\sqrt{2}$

$\Rightarrow MQ=\sqrt{2}$

$\Rightarrow D(1-\sqrt{2},\sqrt{2})$

Orthocentre(centroid) would divide OD in $2:1$

$\Rightarrow G(\frac{2(1-\sqrt{2})}{3},\frac{\sqrt{2}}{3})$.