Question

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

• A. $\frac{2}{\sqrt{10}}$
• B. $\frac{1}{\sqrt{10}}$
• C. $\frac{4}{\sqrt{10}}$
• D. None of these

Answer 1

Answer: (A)

$\frac{2}{\sqrt{10}}$

Answer 2

$9x^{2}-6xy+y^{2}+18x-6y+8=0$ $\Rightarrow \left(\left(3x^{2}\right)-2\times \left(3x\right)\times y+y^{2}\right)+6\left(3x-y\right)+8=0$ $\Rightarrow \left(3x-y\right)^{2}+6\left(3x-y\right)+8=0$ Let $3x - y = z$ $\therefore z^{2}+6z+8=0$ $\Rightarrow z^{2}+4z+2z+8=0$ $\Rightarrow z\left(z+4\right)+2\left(z+4\right)=0$ $\Rightarrow \left(z + 2\right)\left(z + 4\right) = 0$ $\Rightarrow z=-2, z=-4$ $3x-y+2=0\,...\left(i\right)$ or $3x - y + 4 = 0$ If $P_{1}$ be the distance of line $\left(i\right)$ from the origin, then $P_{1}=\frac{2}{\sqrt{9+1}}=\frac{2}{\sqrt{10}}$ Also, if $P_{2}$ be the distance of line \left(ii\right) from the origin,then $P_{2}=\frac{4}{\sqrt{10}}$ So, distance between lines $P=P_{2}-P_{1}=\frac{4}{\sqrt{10}}-\frac{2}{\sqrt{10}}=\frac{2}{\sqrt{10}}$