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.

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