The distance between the pair of parallel lin | vedclass

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(B) The given equation is $x^2+2xy+y^2-8ax-8ay-9a^2=0$. Comparing this with the general second-degree equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,we get $A=

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$5 \sqrt{2} a$

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(B) The given equation is $x^2+2xy+y^2-8ax-8ay-9a^2=0$. Comparing this with the general second-degree equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,we get $A=1, B=1, H=1, G=-4a, F=-4a, C=-9a^2$. The distance $d$ between two parallel lines represented by $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$ is given by the formula $d = 2 \sqrt{\frac{G^2-AC}{A(A+B)}}$. Substituting the values: $d = 2 \sqrt{\frac{(-4a)^2 - (1)(-9a^2)}{1(1+1)}}$ $d = 2 \sqrt{\frac{16a^2 + 9a^2}{2}}$ $d = 2 \sqrt{\frac{25a^2}{2}} = 2 	imes \frac{5a}{\sqrt{2}} = 5\sqrt{2}a$. Thus,the distance is $5\sqrt{2}a$ units.

The distance between the pair of parallel lines represented by $x^2+2xy+y^2-8ax-8ay-9a^2=0$ is $...$ units.

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