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

algebra-precalculusanalytic geometryrotationssoft-question

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= (0,b)$.

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The question is to find the equation of $D'$ such that $D'$ is above $D$ and the distance from any point of $D'$ to $D$ is , say , $4$ units, meaning that the distance between the 2 lines is constant.

I thought of this :

(1) The angle between $D$ and the X axis is $ \alpha = \arctan (a)$.

(2) Let $Q$ be the point where $D'$ crosses the Y axis, and let $R$ be the perpendicular projection of $Q$ on $D$ , we have a right triangle QRP . The angle $\beta = \angle {QPR}= \pi/2 – \alpha$.

(3) The distance between $Q$ and $R$ has to be $4$ ( by the givens of the problem) , so, the relation must hold in $\triangle {PQR}$ : $ PQ \sin(\beta) = 4\space \space (= QR)$, implying that $PQ = 4/ \sin (\beta)$

(4) The difference of height between $D$ and $D'$ is therefore $ c = PQ = 4/ \sin (\beta)$.

(5) From this I conclude that the equation of $D'$ is :

$ y = ax+b +c = ax+b+ \frac {4} {\sin (\arctan (a))}$.

Apparently, that works : https://www.desmos.com/calculator/x0haqu0a0a.

But what are the limitations of this method? For example, what if it were required $D$ to rotate about an arbitrary point ? Also, is there a quicker way to derive the equation of $D'$?

Best Answer

  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.

  2. The formula for the distance between these two parallel lines is $$d = \frac{|b_1-b_2|}{\sqrt{m^2+1}}.$$ You can see the derivation on Wikipedia distance between parallel lines.

  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 $b_2$. $$b_2 = b_1 \pm d\sqrt{m^2+1} $$

    There are two solutions because the second line can be "above" or "below" the first line.

  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 $x = b_1 \pm d$.

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