Question

The distance between the parallel lines $9x^2-6xy+y^2+18x-6y+8=0$ is?

• A. $\frac{1}{\sqrt{10}}$
• B. $\frac{2}{\sqrt{10}}$
• C. $\frac{4}{\sqrt{10}}$
• D. $\sqrt{10}$

Answer 1

The correct option is B $\frac{2}{\sqrt{10}}$

$S\equiv 9x^2-6xy+y^2+18x-6y+8=0$

This equation is of the form :

$S\equiv a{x}^2+2hxy+b{y}^2+2gx+2fy+C=0$ _____ (A)

Reverting the main equation

$S\equiv 9x^2+2(-3)xy+\left(1\right)y^2+2\left(9\right)x+2(-3)y+8=0$

Compairing this equation with equation (A), we get

$a=9;b=1;c=8;h=-3;g=9;f=-3$

Distance between two parallel lines is given by :

$d=2\sqrt{\frac{g^2-ac}{a(a+b)}}$ or $d=2\sqrt{\frac{f^2-bc}{b(a+b)}}$

So, $d=2\sqrt{\frac{9^2-ac}{a(a+b)}}=2\sqrt{\frac{(9{)}^2-9(8)}{a(a+1)}}$

$=2\sqrt{\frac{81-72}{9\times 10}}=2\sqrt{\frac{9}{9\times 10}}$

$\Rightarrow d=\frac{2}{\sqrt{10}}$