Question

Distance between the pair of lines represented by the equation $x^2-6xy+9y^2+3x-9y-4=0$ is

• A. $\frac{5\sqrt{10}}{10}$
• B. $\frac{1}{2}$
• C. $\frac{5}{2}$
  
• D. $\frac{1}{10}$

Answer 1

The correct option is A $\frac{5\sqrt{10}}{10}$

Given Eq

$x^2-6xy+9y^2+3x-9y-4=0$

$x=\frac{-(3-6y)\pm \sqrt{9-36y+36y^2-4(9y^2-9y-4)}}{2}$

$2x=-(3-6y)\pm \sqrt{9-36y+36y^2-36y^2-36y+16}$

$2x=-(3-6y)\pm \sqrt{25}$

$2x=-(3-6y)\pm 5$

$2x=-3+6y+5$ and $2x=-3+6y-5$

$2x-6y=2$ and $2x-6y=-8$

$x-3y-1=0$ and $x-3y+4=0$

Here lines are parallel Distance between $x-3y-1=0$ and $x-3y+4=0$ by formula is

$distance=\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|$

$distance=\left|\frac{-1-4}{\sqrt{1^2+(-3{)}^2}}\right|$

$distance=\left|\frac{-5}{\sqrt{1+9}}\right|$

$distance=\frac{5}{\sqrt{10}}$

$distance=\frac{5\sqrt{10}}{10}$