Question

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

Answer 1

2

Answer 2

Let $Z = x+7y$. The given equation becomes a quadratic equation in $Z$: $Z^2 + 4\sqrt{2}Z - 42 = 0$

Solving this quadratic equation for $Z$ using the quadratic formula $Z = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $Z = \frac{-4\sqrt{2} \pm \sqrt{(4\sqrt{2})^2 - 4(1)(-42)}}{2(1)}$ $Z = \frac{-4\sqrt{2} \pm \sqrt{32 + 168}}{2}$ $Z = \frac{-4\sqrt{2} \pm \sqrt{200}}{2}$ $Z = \frac{-4\sqrt{2} \pm 10\sqrt{2}}{2}$

This gives two distinct values for $Z$: $Z_1 = \frac{-4\sqrt{2} + 10\sqrt{2}}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$ $Z_2 = \frac{-4\sqrt{2} - 10\sqrt{2}}{2} = \frac{-14\sqrt{2}}{2} = -7\sqrt{2}$

Substituting back $Z = x+7y$, we obtain the equations of two parallel lines: Line 1: $x+7y = 3\sqrt{2} \implies x+7y - 3\sqrt{2} = 0$ Line 2: $x+7y = -7\sqrt{2} \implies x+7y + 7\sqrt{2} = 0$

The distance between two parallel lines of the form $ax+by+c_1=0$ and $ax+by+c_2=0$ is given by the formula: $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$

For our lines, $a=1$, $b=7$, $c_1=-3\sqrt{2}$, and $c_2=7\sqrt{2}$. $d = \frac{|-3\sqrt{2} - 7\sqrt{2}|}{\sqrt{1^2 + 7^2}}$ $d = \frac{|-10\sqrt{2}|}{\sqrt{1 + 49}}$ $d = \frac{10\sqrt{2}}{\sqrt{50}}$ $d = \frac{10\sqrt{2}}{5\sqrt{2}}$ $d = 2$