The distance between the parallel lines repre | vedclass

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(D) The given equation is $x^2+2xy+y^2-8mx-8my-9m^2=0$.This can be written as $(x+y)^2-8m(x+y)-9m^2=0$.Let $X = x+y$. Then the equation becomes $X^2-8

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$5 \sqrt{2} m$

Vedclass · Answer

(D) The given equation is $x^2+2xy+y^2-8mx-8my-9m^2=0$.This can be written as $(x+y)^2-8m(x+y)-9m^2=0$.Let $X = x+y$. Then the equation becomes $X^2-8mX-9m^2=0$.Factoring the quadratic equation: $(X-9m)(X+m)=0$.So,the two lines are $x+y-9m=0$ and $x+y+m=0$.These are parallel lines of the form $Ax+By+C_1=0$ and $Ax+By+C_2=0$.The distance between them is given by $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.Here,$A=1, B=1, C_1=-9m, C_2=m$.$d = \frac{|-9m-m|}{\sqrt{1^2+1^2}} = \frac{|-10m|}{\sqrt{2}} = \frac{10|m|}{\sqrt{2}} = 5\sqrt{2}|m|$.Assuming $m>0$,the distance is $5\sqrt{2}m$.

The distance between the parallel lines represented by the equation $x^2+2xy+y^2-8mx-8my-9m^2=0$ is

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