Question

Assertion :The distance between the lines $x^2+6xy+9y^2+4x+12y-5=0$ is $\frac{6}{\sqrt{10}}$ Reason: Distance between the lines $ax+by+c_1=0$ & $ax+by+c_2=0$ is given by $\frac{|c_1-c_2|}{\sqrt{(a^2+b^2)}}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$x^2+6xy+9y^2+4x+12y-5=0.....\left(1\right)$

Consider the homogenous part equal to zero to find the lines

$\therefore x^2+6xy+9y^2=0\Rightarrow (x+3y{)}^2=0$

$\therefore$ required line will be $x+3y+c_1=0,x+3y+2_2=0$

$\therefore (x+3y+c_1)(x+3y+c-2)$ $=x^2+6xy+9y^2+4x+12y-5=0$

$\Rightarrow (x+3y)(x+3y)+c_2(x+3y)+c_1{c}_2=x^2+6xy+9y^2+4x+12y-5$

$\Rightarrow x(c_2+c_1)+y(3c_2+3c_2)+c_1{c}_2=4x+12y-5$

$\Rightarrow c_1+c_2=4,c_1{c}_2=-5$

$c_1=-1,c_2=5$ or $c_1=-5,c_2=-1$

$\therefore$ Required lines are $x+3y-1=0$

& $x+3y+5=0$

$\therefore$ lines $x+3y-1=0$ & $x+3y+5=0$ are parallel.

$\therefore$ Distance between them $=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}=\frac{6}{\sqrt{10}}$

Hence, Assertion (A) & Reason (R) both are correct & Reason (R) is correct explanation for Assertion (A).