Position of a point with respect to parallel lines.

Question

Prove the fact that if a point in the X-Y plane $(x_1,y_1)$ lies between two parallel lines $ax+by+r=0$ and $ax+by+s=0$ then:

$(ax_1+by_1+r)(ax_1+by_1+s)<0$

I tried to assert the above by using the approach that proves the position of points with respect to a certain line. However, I couldn't draw out any conclusive results. Any help is appreciated.

Answer 1

The distance from $(x_1,y_1)$ to the first line is:

$$d_1 = \frac{|ax_1 + by_1 + r|}{\sqrt{a^2 + b^2}}$$

and that to the second line is:

$$d_2 = \frac{|ax_1 + by_1 + s|}{\sqrt{a^2 + b^2}}$$

Since $(x_1,y_1)$ lies between the two lines, we must have that $d_1 + d_2$ is the same as the distance $d$ between the two lines, which is given by

$$d = \frac{|r-s|}{\sqrt{a^2 + b^2}}$$

This implies that $|ax_1 + by_1 + r| + |ax_1 + by_1 + s| = |r-s|$, which obviously only happens if $ax_1 + by_1 + r$ and $ax_1 + by_1 + s$ have opposite signs: if both are non-negative, we get $ax_1 + by_1 + s = 0$, meaning that $(x_1,y_1)$ lies on second line, a contradiction, and if both are non-positive, we get $ax_1 + by_1 + r = 0$, meaning that $(x_1, y_1)$ lies on the first line, a contradiction.

Answer 2

The set of points lying between the two parallel lines is $$\begin{align}\{t(x_r,y_r)+(1-t)(x_s,y_s)&\mid ax_r+by_r+r=0,ax_s+by_s+s=0,t\in(0,1)\}\\=&\{(x,y)\mid \exists u\in(r,s), ax+by+u=0\}\\=&\{(x,y)\mid-ax-by\in(r,s)\}\\=&\{(x,y)\mid(ax+by+r)(ax+by+s)<0\}.\end{align}$$

Answer 3

Using position of point with respect to line,assuming you know it comes from section formula,there can be two cases for lines,Let us call these two lines $L_1$ and $L_2$

There can be 4 cases

1)$S_1<0$,$S_2<0$

2)$S_1>0$,$S_2>0$

3)$S_1<0$,$S_2>0$

4)$S_1>0$,$S_2<0$

where $S_1$ and $S_2$ denote $(ax_1+by_1+r)$ and $(ax_2+by_2+s)$ respectively

from theory of Position of point with respect to line,case 1 and Case 2 indicate that the point lies on same side of both lines,which indicates it cannot be between the two lines

however,

Case 3 indicates That The point lies between origin And $L_2$ and opposite to side of Origin for $L_1$

Case 4 indicates That The point lies between origin And $L_1$ and opposite to side of Origin for $L_2$

both case 3 and 4 hence conclusively bound point to be between line $L_1$ and line $L_2$

Hence combining inequalities in case 3 and case 4

we get $S_1S_2<0$, Which is essentialy what needs to be proved