Question

The length of the perpendicular(s) from the origin to the line passing through $(2,2)$ and perpendicular to the lines$x^2+2xy-3y^2=0$, is/are

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

Answer 1

Answer: (A), (C)

$2\sqrt{2}$ units

Given curve: $x^2+2xy-3y^2=0$
$\Rightarrow x^2+3xy-xy-3y^2=0$
$\Rightarrow (x+3y)(x-y)=0$
So, lines will be
$x+3y=0$ or $x-y=0$
Now
Equation of normal to $x=y$ and passes through $(2,2)$ will be $x+y=4$
so distance from the origin $=\frac{|-4|}{\sqrt{2}}=2\sqrt{2}$ units
Now for line $x+3y=0$ the perpendicular line will be $3x-y=4$
so distance from the origin $=\frac{|-4|}{\sqrt{10}}=\frac{2\sqrt{2}}{\sqrt{5}}$ units