Question

One vertex of the equilateral triangle with centroid at the origin and one side as $x+y-2=0$ is?

• A. $(-1,-1)$
• B. $(2,2)$
• C. $(-2,-2)$
• D. None of these

Answer 1

The correct option is C $(-2,-2)$

$letBCbex+y-2=0$

$AO:OD=2:1,$

$Also,\because △ABD\cong △ADC,\therefore \angle ADB=\angle ADC={90}^{\circ }$

$Fromsectionformula$

$0=\frac{2\alpha +h}{2+1}$

$⟹2\alpha +h=0-\left(4\right)$

$and0=\frac{2\beta +k}{2+1}$

$⟹2\beta +k=0-\left(5\right)$

$or2\alpha +h=2\alpha +k$

$⟹2(\alpha -\beta )=k-h-\left(1\right)$

$\because BCpassesthroughD$

$⟹\alpha +\beta -2=0-\left(2\right)$

$slopeofAD=\frac{\beta -0}{\alpha -0}=\frac{\beta }{\alpha }$

$orslopeofAD=\frac{k-0}{h-0}=\frac{k}{h}$

$SlopeofBCis-1$

$\because y=-x+2$

$⟹\frac{k}{h}\times -1=-1(\because AD\&BCareperpendicular)$

$⟹k=h-\left(3\right)$

$solving\left(1\right),\left(2\right)\&\left(3\right)$

$\left(2\right)⟹\alpha +\beta =2$

$\left(1\right)⟹2(\alpha -\beta )=k-h$

$⟹\alpha -\beta =0$

$and\alpha +\beta =2$

$⟹2\alpha =2$

$⟹\alpha =1$

$\beta =2-\alpha =2-1=1$

$Substituting\alpha \&\beta in\left(4\right)\&\left(5\right)$

$⟹2\alpha +h=0$

$⟹h=-2\alpha =-2$

$2\beta +k=0$

$⟹k=-2\beta =-2$

$\therefore Co-ordinatesofA(-2,-2)$