Question

Assertion :If $x^2+6xy+9y^2+4x+12y-5=0$ represents two parallel lines, then distance between them is $\frac{6}{\sqrt{10}}$ Reason: If $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ represents a pair of straight lines, then the distance between them is given by $\sqrt{\frac{g^2-ac}{a(a+b)}}$

• 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 C Assertion is correct but Reason is incorrect

Let $lx+my+c_1=0$ & $lx+my+c_2=0$

be two parallel lines represented by the given equation then we have

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=(lx+my+c_1)(lx+my+c_2)$

On comparing

$a=l^2,b=m^2,2h=2lm,l(c_1+c_2)=2g$

and $m(c_1+c_2)=2f,c_1{c}_2=c$
Now distance between the parallel lines

$=\frac{|c_1-c_2|}{\sqrt{l^2+m^2}}$ $=\sqrt{\frac{(c_1+c_2{)}^2-4c_1{c}_2}{l^2+m^2}}$

$=\sqrt{\frac{{\left(\frac{2g}{l}\right)}^2-4c}{a+b}}=\sqrt{\frac{\frac{4g^2}{a}-4c}{a+b}}=2\sqrt{\frac{g^2-ac}{a(a+b)}}$

i. e. Reason (R) is false

Again $x^2+6xy+9y^2+4x+12y-5=0$

are two parallel lines given by $(x+3y-1)(x+3y+5)=0$

i.e. $x+3y-1=0$ & $x+3y+5=0$

and distance between them

$=2\sqrt{\frac{g^2-ac}{a(a+b)}}=2\sqrt{\frac{4+5}{1(1+9)}}=\frac{6}{\sqrt{10}}$

so the Assenion (A) is true As Assertion (A) is true & Reason (R) is false,