# Two parallel lines with varying slopes and constant distance ( say 4 units) between them. Let D be

Two parallel lines with varying slopes and constant distance ( say 4 units) between them.
Let D be a line with equation $y=ax+b$, with a a varying number.
So D has a changing slope and is rotating about point $P=\left(0,b\right)$ .
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

eurgylchnj
Step 1
If two lines are parallel, they have the form $y=mx+{b}_{1}$ and $y=mx+{b}_{2}$ . That is, their slopes are the same and their intercepts are different.
Step 2
The formula for the distance between these two parallel lines is
$d=\frac{|{b}_{1}-{b}_{2}|}{\sqrt{{m}^{2}+1}}.$
Step 3
So if you know ${b}_{1}$ and have a particular d in mind, you can rearrange this formula to solve for the second line's parameter ${b}_{2}$ .
${b}_{2}={b}_{1}±d\sqrt{{m}^{2}+1}$
There are two solutions because the second line can be "above" or "below" the first line.
Step 4
In the special case that the lines are vertical, the slope m is infinite so this formula doesn't work. Instead, our lines are simply $x={b}_{1}$ and $x={b}_{1}±d$

Joel French
Step 1
Using vector geometry, 4-or, more generally, d-equals the scalar rejection of $\stackrel{\to }{PQ}$ from the direction vector v of D. So,
$\begin{array}{rl}d& =|\stackrel{\to }{PQ}×\stackrel{^}{\mathbf{v}}|\\ & =\frac{1}{\sqrt{{a}^{2}+1}}|\left(\begin{array}{c}0\\ ±PQ\\ 0\end{array}\right)×\left(\begin{array}{c}1\\ a\\ 0\end{array}\right)|\\ & =\frac{1}{\sqrt{{a}^{2}+1}}|\left(\begin{array}{c}0\\ 0\\ \mp PQ\end{array}\right)|\\ & =\frac{PQ}{\sqrt{{a}^{2}+1}}\\ PQ& =d\sqrt{{a}^{2}+1}.\end{array}$
Thus, the required equation is
$\begin{array}{rl}y& =ax+b±PQ\\ y& =ax+b±d\sqrt{{a}^{2}+1}.\end{array}$