Proportionality between linear segments

Question

This may seem a rather elementary question, but I was wondering if there is a way of, having two linear segments $y_1$ and $y_2$ in accordance to $y=mx+b$ and with different values for $b_1$ and $b_2$, finding a value $c$ so that $y_1=\frac{c}{y_2}$?

I've tried $c=\frac{y_1}{y_2}$ in Matlab, but the scalar returned does not verify $y_1=\frac{c}{y_2}$.

Answer 1

I try to "translate" the request in the OP.

• Case 1: lines with same slope

The following equations $$y_1=m x+ b_1 $$ $$y_2=m x+b_2 $$

describe parallel lines in the plane, with different intercepts $b_1\neq b_2$. Let $A_1,B_1$ be points on $y_1$ and $A_2,B_2$ be points on $y_2$, pairwise distinct.

If we identify two segments $A_1B_1$ resp. $A_2B_2$ on $y_1$, resp. $y_2$, then we can search for the scalar (if it exists) $c\in\mathbb R$ s.t.

$$A_1B_1=\frac{c}{A_2B_2}; $$

We arrive at the solution

$$c=\sqrt{\left((x_1-x_2)^2+(y_1-y_2)^2\right)\left((x_3-x_4)^2+(y_3-y_4)^2\right)}=(1+m^2)|(x_1-x_2)(x_3-x_4)| $$

when $A_1=(x_1,mx_1+b_1)$, $B_1=(x_2,mx_2+b_2)$, $A_2=(x_3,mx_3+b_2)$ and $B_2=(x_4,m x_4+b_2)$.

Searching for a $c\in\mathbb R$ s.t. $y_1=\frac{c}{y_2} $ is a ill-posed problem, instead. One could solve such equation w.r.t. $x$ for all $c\in\mathbb R$, leading to a completely different discussion, though.

• Case 2: lines with different slope

If we consider the case of 2 generic lines, i.e.

$$y_1=m_1 x+ b_1 $$ $$y_2=m_2 x+b_2 $$

and we keep the same notation for the segments $A_1B_1$ resp. $A_2B_2$ on $y_1$, resp. $y_2$ as in Case 1. above, we arrive at the equation

$$A_1B_1=\frac{c}{A_2B_2}$$

which admits solution

$$c=\sqrt{\left((x_1-x_2)^2+(y_1-y_2)^2\right)\left((x_3-x_4)^2+(y_3-y_4)^2\right)}=\sqrt{(1+m_1^2)(1+m_2^2)} |(x_1-x_2)(x_3-x_4)| $$

where$A_1=(x_1,m_1x_1+b_1)$, $B_1=(x_2,m_1x_2+b_2)$, $A_2=(x_3,m_2x_3+b_2)$ and $B_2=(x_4,m_2 x_4+b_2)$.

Answer 2

You can only solve this for all $x$ if $m_1 = m_2 = 0$, so you have two constant lines, or if $c = 0$ and $m_1 = b_1 = 0$.

To see this, notice that you need $$(m_1 x + b_1)(m_2 x + b_2) = c$$ for all $x$. But comparing coefficients, we find $$m_1 m_2 = 0, \qquad m_1 b_2+m_2 b_a = 0, \qquad b_1b_2=c$$ so the first one tells you at least one of the slopes is zero, and then you can use the others to figure out the rest.