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Distance Between a Point and a Line

Date: 08/23/98 at 15:52:32
From: liz
Subject: Equations of straight lines

Find the distance between the origin and the line 3x + 4y = 5.

Date: 08/24/98 at 04:06:05
From: Doctor Pat
Subject: Re: Equations of straight lines


There is a shortcut to get the answer I am going to show you, but it 
really doesn't help you understand without the rest, so here goes.  

The distance from a point to a line is measured, as I'm sure you know, 
along a perpendicular. So first we have to find a line perpendicular 
to 3x + 4y = 5 that passes through the origin. This is easy since we 
know that the slope of a line perpendicular to 3x + 4y = 5 must be the 
negative reciprocal of the slope of 3x + 4y = 5. Since this line has a 
slope of -3/4, the perpendicular must have a slope of +4/3, and since 
it passes through the origin, it has a y-intercept of zero. A line 
with a slope of 4/3 and a y-intercept of zero looks like y = 4/3 x, 
but if we multiply through by three and subtract it can also be 
written as 4x - 3y = 0.  (Hmmmm, compare that to the original line and 
see if you can notice a short cut to find perpendicular lines in 
general form).

Okay, so we know that 4x - 3y = 0 is perpendicular to 3x + 4y = 5. 
Now if we know where these two cross, we can find a point, and then its 
distance from (0,0) is the distance we seek. Well, using systems of 

   3x + 4y = 5
   4x - 3y = 0  

can be rewritten as:
   9x + 12y = 15
  16x - 12y = 0   

and this gives us:

  25x = 15   or   x = 3/5

Substituting this back into either equation (I'll use the one with a 
zero because that seems easier), 4x - 3y = 0,  so  4(3/5) - 3y = 0. 
Thus, 12/5 = 3y and y = 4/5. The two lines intersect at (3/5, 4/5). 
Now all we need to do is find how far this point is from the origin 
(0,0) and we know the shortest distance from the origin to 3x + 4y = 
5. Using the distance formula or Pythagorean Formula (they really are 
the same, aren't they?) we get sqrt(9/25 + 16/25) = sqrt(25/25) = 
sqrt(1) = 1. So that's the distance, one unit.

Now the shortcut. If you do the above a whole bunch of times, you begin 
to notice that you can get the same result by writing the original 
equation on one side and ignoring the equals zero part (3x + 4y - 5). 
To find how far it is from any point (a,b) plug these values in for 
x and y and evaluate the expression. Here are a couple of examples:

   For (3,2): 3x + 4y - 5 = 3(3) + 4(2) - 5 = 12
   For (5,-1): 3x + 4y - 5 = 3(5) + 4(-1) - 5 = 6

Now we just take the absolute value of this number and divide it by the 
square root of the sum of the squares of the coefficients of x and y. 
That just means take the 3 and 4 in front of x and y and square them, 
add the answers together, and take the square root. In this case, 
3^2 + 4^2 = 25 and sqrt(25) = 5, so we divide by five.  

Now to do the original problem and a few others. How far is 
3x + 4y = 5 from each of these points?

   (0,0): Find 3(0) + 4(0) - 5 = -5. Note that |-5| = 5 and divide by 
          5. Thus 5/5 = 1 unit distance.

   (2,5): Find 3(2) + 4(5) - 5 = 21. Note that |21| = 21 and divide by  
          5. Thus 21/5 = 4.2 units.
   (-4,1): Find 3(-4) + 4(1) - 5 = -13. Note that |-13| = 13 and 
           divide by 5. Thus 13/5 = 2.6 units.

Okay, there you have it, the long way and the short way. I hope this 
helps. The shortcut can be really quick, but if you don't see why it 
works, do some more the other way or you will never remember the 
shortcut. Good luck. 

- Doctor Pat, The Math Forum
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Associated Topics:
High School Basic Algebra
Middle School Algebra

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