I need help finding the points on the line to plug into the L formula.The distance between the two parallel lines $\displaystyle L_1: -4x-8y=7, L_2: -4x-8y=-4$
1. The long way to get the distance between the parallels would be:
- choose an arbitrary point on L1
- calculate the equation of the line perpendicular to L1 and passing through the chosen point
- calculate the coordinates of the point of intersection of the perpendicular line with L2
- calculate the distance between the chosen point and the point of intersection
2. The shortcut:
The equations of both lines are given in normal form. Transform the equations into Hesse normal form by dividing through the length of the normal vector. The difference of the two constants yields the distance between the two parallels.
3. With your question:
$\displaystyle \vec n = (-4, -8)~\implies~|\vec n| = 4\sqrt{5}$
$\displaystyle L_1: \dfrac{-4x-8y}{4\sqrt{5}}=\dfrac7{4\sqrt{5}}, ~~~~ L_2: \dfrac{-4x-8y}{4\sqrt{5}}=\dfrac{-4}{4\sqrt{5}}$
4. Therefore the distance is $\displaystyle d = \dfrac7{4\sqrt{5}} - \dfrac{-4}{4\sqrt{5}} = \dfrac{11}{4 \sqrt{5}}$