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### Distance from a Line to a Point

```Date: 07/10/2003 at 23:09:36
From: Jack
Subject: Find the direct distance from the line to the point

Find the direct distance from the line to the point:

4x-3y-5 = 0 , (7,6)
```

```
Date: 07/17/2003 at 09:53:46
From: Doctor Korsak
Subject: Re: Find the direct distance from the line to the point

Hi Jack,

I am sure that a general formula for the answer to your question can
be found in current school and college textbooks, but you can work
out the solution yourself like this:

The closest distance to a line is along another line intersecting the
first at a 90-degree angle; the term for that is an orthogonal line.
The slope m of an orthogonal line is -1/m, where m is the slope of
the given line. (This works only when m is not 0.)

For the example in your question, the slope of the line

4x - 3y - 5 = 0

is 4/3, since you can rewrite the equation as

y = (4x - 5)/3

= (4/3)x - 5/3.

The orthogonal line you want is therefore

y = -(3/4)x + b,

where b is to be determined by the line having to pass through your
starting point (7,6), which means that

6 = -(3/4)7 + b,
b = 6 + 21/4 = 45/4,

so then the orthogonal line is

y = -(3/4)x + 45/4

or

3x + 4y = 45.

Now you can compute the shortest distance between the given point and
line.  First, you can find where the two lines intersect by solving
the pair of equations of the two lines for x and y.  Then you can
calculate the distance to the point (7,6) and you will have your

There is a more direct way to calculate the distance. You can start at
the point (7,6) and go out along a line in the orthogonal direction
toward (or away from) the given line by moving a point (x,y)

x = 7 + 4s/5, y = 6 - 3s/5

where s is a parameter that moves you along the orthogonal line.

Why did I divide by 5? Well, we want to move along the orthogonal line
by a distance s and then solve for s when we hit the given line, so we
prefer to move by multiplying a unit length segment by a factor s. It
so happens that 3 and 4 are the lengths of the sides of a right
triangle with a long side of length 5, and 3^2 + 4^2 = 5^2, so the
point (3/5, -4/5) forms a unit length line segment from the point
(0,0) in the orthogonal direction we need to travel.

Now we calculate s by requiring our point to lie on the given line:

4x - 3y - 5 = 0

substituting our expressions for x and y in terms of s gives

4(7 + 4s/5) -3(6 - 3s/5) - 5 = 0

and now we can solve for s (the distance) directly.

I hope this response is of value to you.  Feel free to write back if
you require further assistance.

- Doctor Korsak, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Linear Equations

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