The distance between the parallel lines given | vedclass

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(C) Given equation is $(x+7y)^2 + 4\sqrt{2}(x+7y) - 42 = 0$. Let $t = x+7y$. The equation becomes $t^2 + 4\sqrt{2}t - 42 = 0$. Using the quadratic for

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$2$

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(C) Given equation is $(x+7y)^2 + 4\sqrt{2}(x+7y) - 42 = 0$. Let $t = x+7y$. The equation becomes $t^2 + 4\sqrt{2}t - 42 = 0$. Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t = \frac{-4\sqrt{2} \pm \sqrt{(4\sqrt{2})^2 - 4(1)(-42)}}{2} = \frac{-4\sqrt{2} \pm \sqrt{32 + 168}}{2} = \frac{-4\sqrt{2} \pm \sqrt{200}}{2} = \frac{-4\sqrt{2} \pm 10\sqrt{2}}{2}$. Thus,$t = 3\sqrt{2}$ or $t = -7\sqrt{2}$. The two parallel lines are $x+7y - 3\sqrt{2} = 0$ and $x+7y + 7\sqrt{2} = 0$. The distance $d$ between two parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. $d = \frac{|7\sqrt{2} - (-3\sqrt{2})|}{\sqrt{1^2 + 7^2}} = \frac{|10\sqrt{2}|}{\sqrt{50}} = \frac{10\sqrt{2}}{5\sqrt{2}} = 2$.

The distance between the parallel lines given by $(x+7y)^2 + 4\sqrt{2}(x+7y) - 42 = 0$ is

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