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### Using the Distance Formula

```
Date: 05/03/97 at 16:31:30
From: ???
Subject: Finding a point on a line a certain distance from another point

Hi Dr. Math,

Can you help me with this problem?

Given a line going through the points (10,10) and (20,15), find the
coordinates of a third point on the line that is 3 units from the
point (10,10).

Thanks!
```

```
Date: 05/04/97 at 07:49:41
From: Doctor Anthony
Subject: Re: Finding a point on a line a certain distance from another
point

To begin with, we need to find the equation of the line. The slope of
the line is given by (difference of y's)/(difference of x's), which is
equal to:

(15-10)/(20-10)
= 5/10
= 1/2

The equation of the line is:

y-10 = (1/2)(x-10)
y-10 = (1/2)x - 5
y = (1/2)x + 5
2y = x + 10

Let the point which is 3 units from (10,10) be (x,y).

If a point is 3 units from another point, that means that the distance
between the two points is three units. This in turn means that we can
use the distance formula (which comes from the Pythagorean theorem).

The distance formula says that the distance d between two points
(x1,y1) and (x2,y2) is given by d^2 = (x2-x1)^2 + (y2-y1)^2.  This
makes a little more sense when you look at the following diagram:

y-axis
/|\
y2|      /*(x2,y2)
|     / |
|    /  |
|   /   |
y1|  *----|
|   (x1,y1)
|
|__|____|_______\
x1   x2      / x-axis

The distance between the two points (x1,y1) and (x2,y2) is the length
of the hypotenuse of the above right triangle, right?  We use the
Pythagorean theorem to find the length of the hypotenuse of a right
triangle.

The length of the base of this triangle is (x2-x1). You can also
write (x1-x2) because if this happens to be a negative number, it
will be squared in the Pythagorean theorem. The height is (y2-y1),
which again can also be written as (y1-y2). If we call the
hypotenuse d, then from the Pythagorean theorem we can write:

d^2 = (x2-x1)^2 + (y2-y1)^2

In our problem, d = 3 and we want to find the point (x,y) that is 3
units away from (10, 10). So we can write the distance formula as:
(x-10)^2 + (y-10)^2 = 3^2 = 9. We substitute y = (1/2)x + 5 (the
equation of our line) in for y:

(x-10)^2 + (y-10)^2 = 9     Original distance formula
(x-10)^2 + (x/2 + 5 - 10)^2 = 9     Substituting y = (1/2)x + 5
(x-10)^2 + (x/2 - 5)^2 = 9     Simplifying
(x-10)^2 + (x/2 - 10/2)^2 = 9     Finding common denominator
(x-10)^2 + ((x-10)/2)^2 = 9     Simplifying
(x-10)^2 + (1/4)(x-10)^2 = 9     Factoring out (1/2)^2 = 1/4
(5/4)(x-10)^2 = 9     Factoring out (x-10)^2
(x-10)^2 = 36/5
x-10 = +- 6/sqrt(5)
x = 10 +- 6/sqrt(5)
x = 10 +- 6*sqrt(5)/5

We substitute x = 10 + 6*sqrt(5)/5 into the equation for our line:

y = (1/2)x + 5                    Equation for line
= (1/2)(10 + 6*sqrt(5)/5) + 5   Substituting
= 5 + 3sqrt(5)/5 + 5            Simplifying
= 10 + 3sqrt(5)/5

The coordinates of the point that is 3 units from (10,10) are:

[10+6sqrt(5)/5, 10+3qrt(5)/5]

What if we use x = 10 - 6*sqrt(5)/5?  Then we get another point that
is 3 units away from (10,10), but in the opposite direction.  Since
your problem only asks for one point, we don't need to figure out the
coordinates of the other point that is three units away from (10,10).
Can you figure out why there are only two points on the line that are
3 units away from (10,10)?

-Doctors Anthony and Rachel,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra

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