



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 X_{1} and Y_{1} ... 











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 
