Question

If $O,G,H$ be the circumcenter, centroid & orthocentre of triangle $ABC$, then $O,G,H$ are collinear & $G$ divide $OH$ in the ratio $1:2$. In an equilateral triangle, these points coincide and in right-angled $△ABC$ orthocentre is the vertex at which triangle is a right angle.

For given two lines $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ the equations of bisectors of angles between these two lines are given by $\frac{a_{1}x+b_{1}y+c}{\sqrt{a_1^2+b_1^2}}=\pm \frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}}$
for the + ve sign we get the equation of bisector of angle in which origin lies. Also $a_1{b}_2+b_1{b}_2>0$, then origin lies in obtuse angle and for $a_1{a}_2+b_1{b}_2<0$, origin lies in acute angle

On the basis of above passage answer the following question:

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

• A. (-1, -1)
• B. (2, -2)
• C. (-2,2)
• D. (-2, -2).

Answer 1

Answer: (D)

(-2, -2).

Given triangle is equilateral & centroid $G(0,0)$ i.e. $O,G,H$ Coincide. Also AD(Median) is its altitude & $G$ divide it in
the ratio $2:1$

Let $D(\alpha ,\beta )$ lies on $BC$

$\therefore x+y-2=0$

$\Rightarrow \alpha +\beta -2=0$

Let $A\equiv (h,k)$ then

$\frac{2a+h}{3}=0,\frac{2\beta +k}{3}=0$

$\Rightarrow h=-2\alpha ,k=-2\beta$ ....(*)

Also $AD$ is $\perp$ to $BC$

$\therefore \frac{h-\alpha }{k-\beta }(-1)=-1$

$\Rightarrow h-\alpha =k-\beta \Rightarrow \alpha =\beta$ using (*)

$\therefore h=-2,k=-2\therefore Ais(-2,-2)$

Hence choice (d) is correct answer.