Teacher2Teacher

Q&A #1895

Teachers' Lounge Discussion: Distance from a point to a line.

_____________________________________
T2T || FAQ || Ask T2T || Teachers' Lounge || Browse || Search || Thanks || About T2T
_____________________________________

View entire discussion
[<< prev]

From: Loyd <loydlin@aol.com>
To: Teacher2Teacher Public Discussion
Date: 2002040503:30:38
Subject: Re: Re: Distance from a point to a line

My previous post had an error in the distance formula on the very last
line.  The post is repeated here with a correction in the last line.  

The problem that is in most Calculus books for finding the distance
 from a point to a line involves a knowledge of vectors.  For those
whose knowledge is limited to algebra, we can find the distance from a
point to a line by the following steps:

1.  The shortest distance will be a perpendicular to the given line. 
This can be found by knowing that if two lines are perpendicular, then
the slopes are the negative reciprocal of each other.  
2.  Once we have the two lines, we can solve them to find the point of
intersection of the two lines.  

3.  Using the given point and the point of intersection, we can then
use the distance formula to find the distance from the given point to
the intersection.  


Suppose we have a point(j,k) and a given line that has the form y=mx+b
.  Find the shortest distance between the point and the line.  This of
course, will be a perpendicular line through the point (j,k).  The
slope of a perpendicular line is -1/m.  This means that the equation
for the perpendicular line is y=(-1/m)x + d.

Substituting the point into the perpendicular line we have: k=(-1/m)j
+ d.  Solving for d, we obtain:  k + j/m = d.  Therefore, the equation
of the perpendicular becomes:

y=(-1/m)x + k+j/m
and the given line is:
y=mx+b

To find the point of intersection, solve for x and y.

y=mx+b
y=(-1/m)x + d

Thus, mx+b=(-1/m)x +d which when solved for x yields:
x=[m(d-b)]/[m^2+1].  Substitute this x value into y=mx+b and we obtain
y=(1/2)[m(d-b)/(m^2+1)]+b


Example:  Given Y=(1/2)x + 4 and a point (5,6).  Find the line
perpendicular to the given line and which passes through the point
(5,6)=(j,k)

y=(-2/1)x + 6+5/(1/2)
y=(-2/1)x + 16

Solve the following pair of equations to find the intersection:
Y=(1/2)x + 4
y=(-2/1)x + 16

Solving the last two equations with the TI-83 will speed the process,
but you can check by setting the last two equations equal and finding
x.  Then you can solve for y.  The resuslt will be: x=4.8 and y= 6.4
or if you prefer, 24/5 and 32/5.

The distance between the two points will be the solution to;

sqrt[(5-4.8)^2 + (6-6.4)^2] which is a little less than .45.  

Post a reply to this message
Post a related public discussion message
Ask Teacher2Teacher a new question


[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || The Math Library || Quick Reference || Math Forum Search
_____________________________________

Teacher2Teacher - T2T ®
© 1994-2014 Drexel University. All rights reserved.
http://mathforum.org/
The Math Forum is a research and educational enterprise of the Drexel School of Education.The Math Forum is a research and educational enterprise of the Drexel University School of Education.