Question

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

• A. $\frac{15}{\sqrt{10}}$
• B. $\frac{1}{2}$
• C. $\sqrt{\frac{5}{2}}$
• D. $\frac{1}{\sqrt{10}}$
• E. $\frac{1}{5}$

Answer 1

Answer: (C)

$x^2-6xy+9y^2+3x-9y-4=0$ ..(i) $\Rightarrow$ $(x-3y+4)(x-3y-1)=0$ $\therefore$ Eq. (i) represented by $x-3y+4=0$ ...(ii) and $x-3y-1=0$ ...(iii) Since, the above line is parallel. $\therefore$ Required distance $=\left|\frac{-4-1}{\sqrt{1+9}}\right|=\frac{5}{\sqrt{10}}$ $=\sqrt{\frac{5}{2}}$