Distance Between 2 Parallel Lines

**thamathkid1729** · March 14th 2011, 06:51 PM

The lines 3

x + 4y = 12 and 3x + 4y = 72 are parallel. Find the distance that separates these lines.

**Prove It** · March 14th 2011, 07:12 PM

What is the difference between the $\displaystyle y$ intercepts?

**earboth** · March 15th 2011, 03:42 AM

Originally Posted by thamathkid1729

The lines 3

x + 4y = 12 and 3x + 4y = 72 are parallel. Find the distance that separates these lines.

I assume that you mean the perpendicular distance between the two parallel lines.

1. Line $l_1: 3x+4y=12$ has the y-intercept $P(0, 3)$ .
Line $l_2: 3x+4y=72$ has the y-intercept $Q(0, 18)$ .

2. The (perpendicular) distance of a point $P(p, q)$ to a straight line with the equation $Ax+By=C$ is calculated by

$d(P,l) = \dfrac{Ap+Bq-C}{\sqrt{A^2+B^2}}$

3. Calculate the distance of P(0, 3) to $l_2$ :

$d(P,l_2)=\dfrac{3 \cdot 0 + 4 \cdot 3 - 72}{5} = -12$

4. So the distance you are looking for has the value 12. The negative sign of d indicates that the point P and the origin are on the same side of the straight line.

**HallsofIvy** · March 15th 2011, 04:31 AM

Originally Posted by thamathkid1729

The lines 3

x + 4y = 12 and 3x + 4y = 72 are parallel. Find the distance that separates these lines.

Oh, blast! I had just claimed that those lines are NOT parallel, then realized that the "3" multiplying x in the first equation, at least on my web viewer, had been separated by a line break!

Here's how I would do that- because I am terrible at memorizing formulas! The slope of the two lines is -3/4 so a line perpendicular to both will have slope 4/3. The equation of a line with slope 4/3 and passing through (0, 3) is $y= \frac{4}{3}x+ 3$ . That will cross the second line where $3x+ 4y= 3x+ 4(\frac{4}{3}x+ 3)$ $= \3x+ \frac{16}{3}x+ 12= 72$ . Then $(3+ \frac{16}{3})x= \frac{25}{3}x= 72- 12= 60$ and $x= \frac{3}{25}60= \frac{36}{5}$ . And, of course, $y= \frac{4}{3}\frac{36}{5}+ 3= \frac{48}{5}+ 3$ .

That is, the line perpendicular to both given lines intersects them at $\left(0, 3\right)$ and $\left(\frac{36}{5}, \frac{48}{5}+ 3\right)$ . I have left the y value in that form because we need to subtract 3 from it to find the distance between those points.

The distance is $\sqrt{\left(\frac{36}{5}\right)^3+ \left(\frac{48}{5}\right)^2}= \sqrt{\frac{3600}{25}}= \sqrt{144}= 12$ .

(Added after thinking it over) However, note Prove It's post. The first line passes through the point (0, 3) and the second through (0, 18). The difference between the two y intercepts is 18- 3= 15. Now that is NOT the distance between the lines (which is always measured perpendicular to both lines) but if we were to drop a perpendicular from, say, (0, 3) to the second line, it is the hypotenuse of a right triangle that has the common perpendicular to both lines as a leg. Further, a little geometry shows that those two legs in the ratio of 3 to 4 since the slope of both lines is -3/4. That is, if the perpendicular distance is "x" then the other leg has length (3/4)x so we must have $x^2+ 9x^2/16= 15^2$ or $\frac{25}{16}x^2= 15^2$ ..
$x^2= \frac{16}{25}15^2}= 144$ .

**bjhopper** · March 15th 2011, 12:09 PM

Without getting involved in a lenghty response this problem can be solved quickly using similar triangles of 3,4,5 right triangles where the one defining the lenght of the perpendicular has a hypot of 15

bjh

**Plato** · March 15th 2011, 12:25 PM

Originally Posted by thamathkid1729

The lines 3

x + 4y = 12 and 3x + 4y = 72 are parallel. Find the distance that separates these lines.

It can be shown that the distance between the parallel lines
$Ax+By+C=0~\&~Ax+By+D=0$ is simply $\dfrac{|C-D|}{\sqrt{A^2+B^2}}$ .

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