Question

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

• A. $15\sqrt{10}$
• B. $\frac{1}{2}$
• C. $\sqrt{\left(\frac{5}{2}\right)}$
• D. $\frac{1}{\sqrt{10}}$

Answer 1

The correct option is C

$\sqrt{\left(\frac{5}{2}\right)}$

Explanation for the correct option:

Finding the distance between the pair of lines:

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

The given equation is of the form $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

Comparing both the equations we get,

$a=1,h=-3,b=9,g=\frac{3}{2},f=-\frac{9}{2},c=-4$

The given equation represents a pair of lines,

hence, $∆=0,h^2=ab,d=2\sqrt{\frac{g^2-ab}{a(a+b)}}$

Sice $∆=0$, then

$$\begin{gathered} abc+2fgh-b{g}^2-a{f}^2-c{h}^2=0 \\ \text{Substituting the values we get} \\ -36+\frac{81}{2}-\frac{81}{2}-\frac{81}{4}+36=0 \\ 0=0 \end{gathered}$$

Thus it satisfies the condition.

Now finding the distance,

$\begin{array}{rcl}d& =& 2\sqrt{\frac{g^2-ab}{a(a+b)}}\\ & =& 2\sqrt{\frac{{\left({\displaystyle \frac{3}{2}}\right)}^2-\left(1\times 9\right)}{1(1+9)}}\\ & =& 2\sqrt{\frac{{\displaystyle \frac{9}{4}}-9}{10}}\\ & =& 2\sqrt{\frac{{\displaystyle \frac{9-36}{4}}}{10}}\\ & =& 2\sqrt{\frac{{\displaystyle 25}}{40}}\\ & =& 2\sqrt{\frac{25}{4\times 10}}\\ & =& \sqrt{\frac{25}{10}}\\ & =& \sqrt{\frac{5}{2}}\end{array}$

The distance between the pair of lines represented by the equations $x^2-6xy+9y^2+3x-9y-4=0$ is $\sqrt{\left(\frac{5}{2}\right)}$

Hence, option (C) is the correct answer.