Question

Distance between the two lines represented by the line pair, $x^2-4xy+4y^2+x-2y-6=0$ is

• A. $\frac{1}{\sqrt{5}}$
• B. $\sqrt{5}$
• C. $2\sqrt{5}$
• D. None

Answer 1

The correct option is B $\sqrt{5}$

This can be done by two methods:

1. Conceptual Method (Good for boards)
2. Formula based method (Good for Competitive Exams)
   

Solving by Method 1-

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

$\Rightarrow (x-2y{)}^2+(x-2y)=6$

Let $(x-2y)=u$

$\Rightarrow u^2+u-6=0$

Now, by applying method of splitting the middle term

$\Rightarrow u^2+3u-2u-6=0$

$\Rightarrow (u+3)(u-2)=0$

$\Rightarrow u=-3$ or $u=2$

$\Rightarrow x-2y=-3$ and $x-2y=2$ are the equation of the two lines represented by the line pair and as the slope of both the lines are equal, hence, the lines are parallel.

The distance between parallel lines is the distance along a line perpendicular to them.

Let $y=m_{1}x$ be the line perpendicular to the pair of lines.

Since, slope of given line $m_2=\frac{1}{2}$

$\Rightarrow m_1=\frac{-1}{m_2}=-2$

So, equation of perpendicular line will be $y=-2x$

Now, line $y=-2x$ will intersect line $x-2y+3=0$ at $P_1\left(\frac{-3}{5},\frac{6}{5}\right)$ and $x-2y-2=0$ at $P_2\left(\frac{2}{5},\frac{-4}{5}\right)$

$\Rightarrow$distance between givel line pair is equal to the distance b/w $P_1$ and $P_2$

$\Rightarrow \sqrt{{(\left(\frac{-3}{5}\right)-\left(\frac{2}{5}\right))}^2+{(\left(\frac{6}{5}\right)-\left(\frac{-4}{5}\right))}^2}$

$\Rightarrow \sqrt{1+4}$

$\Rightarrow \sqrt{5}$

Hence, option $B$ is correct.

Solving by Method 2-

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

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

On comparing, we get

$a=1,h=-2,b=4,g=0.5,f=-1,c=-6$

Now, for lines to be parallel

$h^2=ab$ and $a{f}^2=b{g}^2$

Since,these two equations are satisfied for the above values of $a,h,b,g,f,c$

Hence, the two lines are parallel.

Now applying the formula to calculate the distance b/w parallel pair of lines.

Distance= $2\sqrt{\frac{g^2-ac}{a(a+b)}}$

$\Rightarrow$ distance$=2\sqrt{\frac{{0.5}^2-\left(1\right)(-6)}{1(1+4)}}$

$\Rightarrow distance=\sqrt{5}$

Hence, option $B$ is the answer.