Question

If $A$ is the orthocentre of the triangle formed by $2x^2-y^2=0,x+y-1=0$ and $B$ is the centroid of the triangle formed by $2x^2-5xy+2y^2=0,7x-2y-12=0$, then the distance between $A$ and $B$ is

• A. $\sqrt{5}$
• B. 1
• C. 5
• D. $\sqrt{2}$

Answer 1

Answer: (A)

Given,
$2x^2-y^2=0$
$\Rightarrow y^2=2x^2$
$\Rightarrow y=\pm \sqrt{2}x$ and $x+y=1$

$\because$ Point of intersection of perpendicular is orthocentre.
The two perpendiculars are $y=x$ and $x-\sqrt{2}y+\sqrt{2}-1=0$
$\therefore$ Point of intersection of $y=x$ and $x-\sqrt{2}y+\sqrt{2}-1=0$
$\therefore A=(1,1)$
Now, $2x^2-5xy+2y^2=0$
$\Rightarrow (2x-y)(y-2x)=0$
$\Rightarrow y=2x$ or $x=2y$ and $7x-2y-12=0$

$\therefore$ Centroid $=\left(\frac{0+4+2}{3},\frac{0+8+1}{3}\right)\left(\frac{6}{3},\frac{9}{3}\right)=(2,3)=B$
Now, required distance
$=\sqrt{1+4}=\sqrt{5}$