Find exact perpendicular distance between two parallel lines

**roger1505** · June 16th 2012, 05:18 PM

Find the exact perpendiuclar distance between two parallel lines.
a)y=3x-2 and y=3x+3
please explain your answer

**Plato** · June 16th 2012, 05:54 PM

Originally Posted by roger1505

Find the exact perpendiuclar distance between two parallel lines.
a)y=3x-2 and y=3x+3
please explain your answer

There is a formula for the distance from a point to a line. FIND IT!
Then find a point on one of the parallel line, use that formula to find the distance from that point to the other line.

**roger1505** · June 16th 2012, 06:08 PM

can you please do it. mainly because its a parallel, and i dont know how to sub it in

**Wilmer** · June 16th 2012, 06:32 PM

Your teacher did not show how...or you missed classes?

**roger1505** · June 16th 2012, 06:53 PM

i wasnt in class when they taught this, but does anyone know?

**richard1234** · June 16th 2012, 07:11 PM

You can easily use the "formula" for the distance between a point and a line. For example, you have $y = 3x-2$ so you know that the point (0,-2) is on the line. So use the formula to find the distance between (0,-2) and the line $y = 3x-2$ .

Or, you can skip the formula and draw the graph. Use geometry.

**bjhopper** · June 17th 2012, 07:30 AM

Hi roger1505,

1 graph the lines
2 erect a vertical line @x=1. Apoint (1,1) on line y=3x-2 is produced
3@(1,1) erect a perpendicular to y=3x+3
4 write an equation for the perpendicular
5 solve for intersection of 4 and y=3x +3
6 use distance formula to find question d
d^2 = delta y^2 + delta x^2 (slope diagram between (1,1) and solution of 5 above

There is a simple solution using trig and the slope angle of the parallel lines

**Plato** · June 17th 2012, 08:13 AM

Originally Posted by roger1505

i wasnt in class when they taught this, but does anyone know?

If $Ax+By+C=0$ is a line where $|A|+|B|\ne 0$ and $P(p,q)$ is a point then the distance from that point to the line is $\frac{|Ap+Bq+C|}{\sqrt{A^2+B^2}}~.$

**Soroban** · June 17th 2012, 03:12 PM

Hello, roger1505!

If you are allowed to use Trigonometry, here is another solution.

Find the exact perpendicular distance between two parallel lines: . $\begin{array}{ccc}y &=& 3x-2 \\ y &=& 3x + 3 \end{array}$

Code:

        |
        |   /
        |  /        /
        | /        /
        |/        /
      A *        /
      : |@*     /
      : |   *  /
      : |     * C
      5 |    /
      : + - / - - - - 
      : |  /
      : | /
      : |/ @
      B * - - - - E
        |

The two lines have y- intercepts at 3 and -2, and identical slopes.
$AB \,=\,5.$
Draw $AC$ perpendicular to the line through $B.$

Let $\theta = \angle CBE$
Then the slope of $BC \,=\,\tan\theta \,=\,3$
Note that $\angle CAB \,=\, \theta.$

We have: . $\tan\theta \:=\:\frac{3}{1}\:=\:\frac{\text{opp}}{\text{adj}}$
$\theta$ is in a right triangle with: $opp = 3,\;adj = 1$
Hence: . $hyp \,=\, \sqrt{10} \quad\Rightarrow\quad \cos\theta \,=\,\tfrac{1}{\sqrt{10}}$

In right triangle $ACB\!:\;\;\cos\theta \,=\,\frac{AC}{5} \quad\Rightarrow\quad AC \:=\:5\cos\theta \:=\:5\left(\tfrac{1}{\sqrt{10}}\right)$

Therefore: . $AC \:=\:\frac{\sqrt{10}}{2}$

**Plato** · June 17th 2012, 03:55 PM

Originally Posted by roger1505

Find the exact perpendiuclar distance between two parallel lines.
a)y=3x-2 and y=3x+3
please explain your answer

As long as we are into spoon feeding, look at reply #8.
Rewrite line #1 as $3x-y-2=0$ . Note that $(1,6)$ is on $y=3x+3$ .

Apply the formula from reply #8: $\frac{|3(1)-1(6)-2|}{\sqrt{(3)^2+(-1)^2}}=\frac{5}{\sqrt{10}}=\frac{\sqrt{10}}{2}.$

**louisejane** · August 5th 2012, 11:09 PM

$\frac{|3(1)-1(6)-2|}{\sqrt{(3)^2+(-1)^2}}=\frac{5}{\sqrt{10}}=\frac{\sqrt{10}}{2}.$

This is the best and simplest way of finding the perpendicular distance between two parallel lines. Hope you get it right.

**Wilmer** · August 6th 2012, 05:28 AM

What is your point LJ? Already answered 2 months ago...

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