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

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Aicharya · Accepted Answer

Given equation is $(x+7 y)^2+4 \sqrt{2}(x+7 y)-42=0$ Put $x+7 y=t$ $\begin{aligned} & \Rightarrow \quad t^2+4 \sqrt{2}(t)-42=0 \ & \Rightarrow \quad t=\frac{-4 \sqrt{2} \pm \sqrt{32+168}}{2} \ & =\frac{-4 \sqrt{2} \pm 10 \sqrt{2}}{2} \ & =3 \sqrt{2},-7 \sqrt{2} \ & 	herefore \quad x+7 y=3 \sqrt{2} \ & \end{aligned}$ and $x+7 y=-7 \sqrt{2}$ $\Rightarrow \quad x+7 y-3 \sqrt{2}=0$ and $\quad x+7 y+7 \sqrt{2}=0$ $	herefore$ Distance between parallel lines $=\frac{|7 \sqrt{2}+3 \sqrt{2}|}{\sqrt{1^2+7^2}}=\frac{10 \sqrt{2}}{5 \sqrt{2}}=2$

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

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