Algebra 1 Point To Line Distance
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How Far
Point To Line Distance

 
 Here's a problem that test writers love.
 
 How far is it from the point (1,6) to the line Y = 2X - 1?
 
 How do you solve a crazy thing like this?
 
 First calm down.
 Then draw the graph of the line
 and also stick that other point in there too.
 
 Find any two points on the line.
 Say, where X = 0 and where X = 2 ...
 
When X = 0 When X = 2
   
Y = 2X - 1 Y = 2X - 1
Y = 2(0) - 1 Y = 2(2) - 1
Y = 0 - 1 Y = 4 - 1
Y = - 1 Y = 3
 
 So our two points are (0,-1) and (2,3)
 now we can graph the line (and that other point too.)
 

 
 Now we need to find how far it is
 from the point to the line.
 
 What they actually mean is:
 "How far away is the point (1,6)
 from the closest point on the line?"
 
 To get to that closest point on the line
 we need to draw the shortest line possible
 that connects the point and the line.
 

 
 It turns out that when we do this,
 we are drawing a line that is "perpendicular"
 to the line we have.
 
 Remember that perpendicular means that the two lines
 form 90 degree angles where they cross AND
 

(the slope of one line) (the slope of the other line) = -1

 
 If we know the slope of the line we have
 we can find the slope of the line that is perpendicular.
 
 The question is,
 what is the slope of the line that we have?
 
 The equation is Y = 2X - 1 so the slope is 2.
 Remember that when an equation is written like this,
 the slope is the number next to the X.
 
 So now we can find the slope of the other line ...
 

 
 Now we have the slope of the other line (-1/2)
 and a point on the other line (1,6)
 We can use that nasty equation from two pages ago
 to find the equation of the line.
 

 
 Now we put our point in for X1 and Y1 ...
 

 
 And turn this mess into Y = stuff ...
 

 
 Now we need to figure out the point 
 where the two lines cross.
 Remember how we do that?
 WOW! All of this glop that we've been learning
 is all being used on this one problem!
 (I guess that's why test writers
 like this kind of problem)

 
 At the point where the two lines cross,
 the Y values (and the X values too) in both equations
 are the same number.
 
 Since the Y values are the same,
 we can join the two equations on the Y values ...
 

 
 Solve for X by moving X's to the left
 and numbers to the right.
 

 
 Now find Y using either equation.
 

Y = 2X - 1

Y = 2(3) - 1
Y = 6 - 1

Y = 5

 
 So the point where the two lines cross is 
 X = 3, Y = 5 also known as the point (3,5).
 

 
 Now all that's left to do is to figure out the distance
 between the point that we were given (1,6)
 and the point that we just found (3,5).
 YOU KNOW, EXACTLY THE STUFF 
 THAT WE DID ON THE LAST PAGE!!!
 
 Let's zoom in on the drawing
 so we can see the area we're looking at
 a little bit better ...
 

 
 The distance between the two points is ...
 

 
 So the distance between the point
 and the closest point on the line is about 2.236.
 Again, we don't need to worry about the +/- thing
 we get with roots because the length of this line
 is a positive number.
 

   copyright 2005 Bruce Kirkpatrick

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