The distance between the two parallel lines r | vedclass

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(D) The given equation is $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$. We can rewrite this as $2(4x^2 - 12xy + 9y^2) - 3(2x - 3y) - 5 = 0$. This simplifie

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$\frac{7}{2\sqrt{13}}$

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(D) The given equation is $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$. We can rewrite this as $2(4x^2 - 12xy + 9y^2) - 3(2x - 3y) - 5 = 0$. This simplifies to $2(2x - 3y)^2 - 3(2x - 3y) - 5 = 0$. Let $t = 2x - 3y$. Then the equation becomes $2t^2 - 3t - 5 = 0$. Factoring the quadratic,we get $(2t - 5)(t + 1) = 0$. So,$2x - 3y = 5/2$ and $2x - 3y = -1$. These are two parallel lines of the form $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$,where $A = 2, B = -3$. Rewriting the lines: $2x - 3y - 2.5 = 0$ and $2x - 3y + 1 = 0$. The distance $d$ between parallel lines is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. $d = \frac{|-2.5 - 1|}{\sqrt{2^2 + (-3)^2}} = \frac{|-3.5|}{\sqrt{4 + 9}} = \frac{3.5}{\sqrt{13}} = \frac{7/2}{\sqrt{13}} = \frac{7}{2\sqrt{13}}$.

The distance between the two parallel lines represented by $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$ is

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