Hey, guys,
welcome to Algebra 1.
Today's lesson focuses on
writing the equation of a line
when given two points
on the line.
You've written the equation
of a line given its graph,
given the slope and a point.
This is a spin-off
of what you already know.
You ready to get going?
Let's start.
Okay, so just to jog
your memory first,
you know how to handle
this kind of problem, right?
Given the graph of a line,
you can identify
the Y intercept
and count off that slope.
You can figure out
the equation of the line
and write it in slope
intercept form.
And you've handled this type,
right?
You know given the slope
of a line
and given a point on the line,
you can use point slope form
and process that problem,
and get to the slope intercept
form of the line.
For this kind of problem,
you're going to be given
two points on the line.
There's two steps
in the process.
First, figure out
what's the slope of the line
that passes through those
two points you were given.
Then, after you
have the slope,
use that slope, and one of
the points you were given,
and the point slope form,
and get to the slope
intercept form.
That sounds like a lot,
but it'll make more sense
when I show it to you.
So here's our first example.
We'll write
the equation of the line

that passes through the points
(8, 1) and (10, 2).
First, figure out
the slope of this line.
There's a couple of ways
you can find the slope
of a line:
use graph paper
and plot these two points
and count out that slope,
or use the slope formula
to get it.
I don't have a graph paper,
so I'll use the slope formula
to find the slope
of this line.
I'll label one of these points
X sub-1, Y sub-1.
I'll label the other point
X sub-2, Y sub-2,
and I'm going to label--
this is my first step--
I'll determine the slope.
Okay?
Remember, the slope formula
was M equals Y2
minus Y1 over X2.
Oops--wrote another Y,
let's get that cleaned up.
There we go--
so used to writing Y's--
over X2 minus X1.
There's our formula
for slope.
I'll fill in
what needs to go where
and get the slope of the line
that passes through
these two points.
So Y2 minus Y1
would be 2 minus 1.
Then, X sub-2 minus X sub-1
would be 10 minus 8.
Okay.
And let's clean this up.
So 2 minus 1, that's 1.
10 minus 8,
that's 2, right?
So the slope of the line
that passes through
these two points is one-half.
So I've got my slope.
Now I'm going to use
point slope form.
I've got my slope.
I can use either one
of those points
to get the equation
of this line.
Now, even though
we've labeled this point
X sub-1, Y sub-1,
and this one X sub-2,
Y sub-2,
we could have
done the opposite:
labeled the (10, 2)
with the X sub-1, Y sub-1,
or labeled the (8, 1)
with the X sub-2, Y sub-2.
It doesn't matter, just stay
consistent with what you use.
When we do step two,
it doesn't matter
which point I use
because they're both
two points on my line.
I'll use (8, 1)
because I've already
labeled it X sub-1, Y sub-1.
But if I'd chosen
to use (10, 2),
I'd still have
the same answer.
Just know that
while you're working.
Let's keep going here.
I'm going to label this
step two
and get my point slope form
up here.
Okay.
I've got Y minus Y sub-1.
Let's fill in
our information here.
Let's get some more workspace.
We've got to do
a little moving around first.
Okay.
That's good.
All right, so Y minus--
our Y sub-1 is 1 here,
but like I said, we could be
using this point--equals M,
which is a half,
times X minus--
our X sub-1 is 8.
Okay?
We'll clean this up
and get to that
slope intercept form equation.
I'll bring down the left side.
I'll the distributive property
on the right.
So one-half times X,
that's one-half X.
One-half times negative 8,
I'll come off to the side
over here and do that.
So one-half
times negative 8 over 1.
1 times negative 8
is negative 8.
2 times 1 is 2.
Negative 8 divided by 2,
that's negative 4, okay?
So back over here to my work.
One-half times negative 8
is negative 4, right?
I'm just going
to clean this up.
I need to get Y by itself,
so I'll add 1
to both sides.
So plus 1, plus 1.
That's going to get
wiped out.
So Y equals one-half X.
And negative 4 plus 1,
that's negative 3.
And you got it,
you're all done.
So it's two steps.
It's lengthy, but when
you get the swing of it,
you can do these quickly.
Remember,
we were given two points.
First,
we calculated the slope,
we used the slope formula,
and after we had that slope,
we used point slope form,
and we got to that final
equation of the line.
Let's try another one.
Write the equation
of the line that passes

through the points
(negative 2, 7) and (0, 1).
So what did we do first
on that last one?
Good--
we labeled these points
so that we could get
to that slope, right?
So I'll label this one
X sub-1, Y sub-1.
I'll label this one
X sub-2, Y sub-2.
Then I'll go ahead
and use the slope formula
to calculate the slope.
So M equals Y2 minus Y1
over X2 minus X1.
So Y sub-2
minus Y sub-1,
that would be 1 minus 7
over X sub-2 minus X sub-1,
so 0 minus negative 2.
I've got that
double negative.
I'll handle that.
All right, so 1 minus 7,
that's negative 6.
0 plus 2, that's 2, right?
And negative 6 divided by 2
is negative 3.
Now I know the slope
is negative 3.
Now, we can go ahead and...
good, move on
to step two.
Use your point slope form,
and then we can
get at the final equation.
Let's get point slope form
down first.
Y minus Y sub-1
equals M times X
minus X sub-1.
Scroll down a little bit,
but still enough
to see my point.
It kind of disappeared,
but I can see it.
So Y minus--our Y sub-1
is 7--equals M,
which is negative 3...
X minus--and our X sub-1
is negative 2.
Let's get some more room
up here to work.
We're always moving
something around up here.
So we can bring
the left side down--
that Y minus 7 equals...
Let's handle that double
negative before anything else.
All right, so negative 3
times X,
that's negative 3X.
Negative 3 times
positive 2,
that's negative 6.
Then our final step
on this one?
Good--plus 7, plus 7.
That's going to cancel out,
and we have Y equals--
bring down that negative 3X,
and then negative 6 plus 7
is a positive 1.
You got it, guys,
all done with that one.
Bringing it all together
one more time.
Scroll back up.
We were given two points.
First, we labeled those points
to use the slope formula
and calculate the slope.
Then, after we had
the slope formula,
we used the slope
and one of our given points
and put that
in point slope form,
worked that all out,
and got to our final equation:
Y equals negative 3X plus 1.
Okay?
All right,
give this one a try.
Press pause, take your time,
work our process, and write
the equation of the line
that passes through
these two points.
To check your work,
press play.

(female narrator)
Write the equation of the line
that passes through the points

(4, negative 3)
and (12, negative 5).
All right, let's see
how you did with this one.
I'll start out
by labeling these points.
First thing I'm going to do
is get the slope.
Let's get that formula
written down.
We've got that down.
I've got the formula,
let's plug in our values.
So Y sub-2 minus Y sub-1
will be negative 5
minus negative 3.
So negative 5
minus negative 3
over X sub-2 minus X sub-1,
12 minus 4.
Let's clean up
that double negative situation.
All right,
so negative 5 plus 3,
that's negative 2.
12 minus 4, that's 8.
I can reduce this fraction;
it's not in simplest form yet.
If I divided the numerator
and the denominator by 2,
I could reduce this
to negative 1/4th, right?
So the slope of this line
is negative 1/4th.
Now, let's jump into
point slope form.
So Y minus Y sub-1
equals M times X
minus X sub-1.
Okay?
Scroll a little more.
Hopefully, we can still
see the point.
So Y minus
negative 3 equals--
our M is negative 1/4th,
parentheses X minus--
and our X sub-1 is 4.
Scroll down
and get some more workspace.
All right?
Clean up that
double negative.
So on the left side,
we have Y plus 3 equals--
and, on the right,
we have negative 1/4th.
Oop--I'm getting ahead
of myself here.
We need to apply
the distributive property,
get that right side
under control.
Okay, so negative 1/4th
times X,
that's negative 1/4th X, okay?
Now, negative 1/4th
times negative 4--
let's do some scratch work
over here on the right.
So negative 1/4th
times negative 4.
And I know I can put that
negative 4 over 1.
So negative 1 times negative 4
is positive 4.
4 times 1 is 4,
and 4 divided by 4 is 1.
Right?
Back to our problem over here,
negative 1/4th times negative 4
is a positive 1.
Then, the last step,
subtract 3 from both sides.
Now, this is going
to cancel out over here,
and we have Y equals...
I'll bring down
that negative 1/4th X,
and 1 minus 3,
that's negative 2.
And you got it, guys.
All done with that one.
Now, before we leave this,
I'm always going through
those special lines,
and I'm going to show you
how to interpret them
so you can quickly
get to the answer.
"Write the equation
of the line

that passes through the points
(negative 3, 4) and (5, 4)."
Okay.
There's a couple ways
to attack this.
You might look at the points
and initially notice that
they have the same Y coordinate.
That's a hint to you.
If the Y's are constant,
they didn't change,
then that means I have
a horizontal line.
You could spin off
from there and say,
"Okay, if this line
is horizontal,
I'm only concerned
with those Y values."
They're both 4.
So the equation of the line
is Y equals 4.
That's one way
to attack this.
And you should
get in the habit of,
every time you're given
two points, check and see:
Do they have constant Y's
or constant X's?
Is something not changing
with those two points?
If that doesn't
immediately jump out at you,
this is how you'll end up
at the equation
of that special line.
You'll go through
your normal process
of first getting the slope.
Let's erase here so we can get
the workspace going.
If that didn't initially
jump out at you,
then go through your process.
Label X sub-1, Y sub-1.
X sub-2, Y sub-2.
Calculate your slope.
This is step one.
So M equals Y sub-2
minus Y sub-1
over X sub-2 minus X sub-1.
Okay, so Y sub-2
minus Y sub-1,
that'd be 4 minus 4...
over X sub-2 minus X sub-1:
5 minus negative 3, right?
You got that
double negative situation
in your denominator,
so handle that.
So 4 minus 4 is 0.
5 plus 3 is 8.
0 divided by 8 is 0.
So that means this line
that passes through
these two points,
has a 0 slope.
And if it has a 0 slope,
it has to be horizontal.
So if this
is a horizontal line,
then you're only concerned
with the Y values
because it's the Y values
that stay constant
on a horizontal line.
At that point, you could
jump to your answer.
The equation of this line
has to be,
because those Y's are constant,
that Y equals 4.
And you can get it
that way, okay?
If you don't initially see--
when the points
are given to you,
if it doesn't jump out
that the Y values are constant,
go through your process.
You'll see, at the end,
the slope will be 0.
Then think, "Well,
if this slope is 0,
"then this line is horizontal,
and I'm only concerned
with the Y values here."
Then get your answer,
which in this case
would be Y equals 4.
Now give this one a try.
Actually, let me work
this one out with you.
I see here that we're given
the points

(1, negative 6) and (1, 2).
We'll write
the equation of the line
that passes through
these two points.
If you look at the points,
notice that they both
have an X coordinate of 1.
So the X coordinates
stayed constant.
And if the X coordinates
stayed constant,
then that means this line
has to be vertical.
And if this
is a vertical line,
then I'm only concerned
with the X values.
And if they're both 1,
then the equation of this line
is X equals 1.
You can be done that fast.
That's only if you notice,
when you start,
that these equations
have the same value for X.
If that doesn't immediately jump
out at you, that's okay.
We'll go through our process,
and it'll jump out at you.
Let me show you
what I mean.
Okay. Let's get my pen back.
Here we go.
I'll go through my process.
I'll label the points
X sub-1, Y sub-1,
X sub-2, Y sub-2.
Okay, so step one,
let's get that slope.
So M equals Y sub-2
minus Y sub-1
over X sub-2 minus X sub-1,
all right?
So Y sub-2 minus Y sub-1
would be 2
minus negative 6
over X sub-2 minus X sub-1,
would be 1 minus 1.
I'll handle this
double negative first.
Okay, 2 plus 6, that's 8.
1 minus 1, that's 0.
Here you'll see it.
You cannot divide
a number by 0--
that's undefined.
You can't have
a numerical value
and then divide it into
zero parts, it's not possible.
In a situation where you have
0 in the denominator,
you are in a situation
where your slope,
or your value, is undefined,
right?
In this case, because we are
calculating slope,
we know that the slope
of the line
that passes through these
two points is undefined.
So then, again,
we get to thinking, right,
if the slope of this line
is undefined,
then this line is vertical.
It has to run
just straight up and down.
If this line is vertical,
then I'm only concerned
with the X's.
The X's are both 1.
So the equation
of my line is X equals 1.
And you're done, okay?
Again, that's another way
to get at the equations
of these special lines.
If it doesn't jump out
when you see the points,
go through your process.
If at the end
of step one, you see
your slope is undefined,
you have a vertical line
and the equation of that line
is only concerned
with those X values.
So in this case,
it was X equals 1.
Okay? Let's keep going.
It's time for you
to try a couple.
You knew it was coming,
so press pause.
Take a few minutes and work
through these problems.
See if it jumps out or
if you go through the process.
Either one is fine.
To check your answers,
press play.

(female narrator)
Write the equation of the line
that passes through
the given points.
Letter A: (8, 1)
and (8, negative 5).

Letter B:
(2, negative 3)
and (4, negative 3).
All right,
let's see how you did.
So for A, the equation
of the line
that passes through (8, 1)
and (8, negative 5)
is X equals 8.
For B, the equation of the line
that passes through

(2, negative 3)
and (4, negative 3)
is Y equals negative 3.
Now, to see how I got these,
I'll show you.
Here we go.
Like I said,
it may have
jumped out at you here
that your X coordinates
were constant,
so that this had to be
a vertical line.
If it didn't, then you went
through the process.
So label the points--
let's get that pen back--
X sub-1, Y sub-1;
X sub-2, Y sub-2.
Get your slope formula
written down.
Now, fill in your values.
So Y sub-2 minus Y sub-1
would be negative 5 minus 1
over X sub-2 minus X sub-1,
would be 8 minus 8.
Okay? So negative 5 minus 1,
that's negative 6.
8 minus 8, that's 0.
Again, I see 0
in the denominator.
It's not possible
to divide a number by 0,
so this slope is undefined.
And if I have a line
with an undefined slope,
then that means I have
a vertical line.
And if I have
a vertical line,
then I'm only concerned
with these X's.
These X's are both 8,
so the equation of my line:
X equals 8,
and you got it.
You're all done with that one.
Let me show you
how I got the next one.
Either it jumped out
that the Y values
were both negative 3,
so this was a horizontal line,
making the equation
Y equals negative 3,
or you went through
the process.
So label your points.
Okay?
Now, go ahead
and get your slope.
Let me get that slope formula
up there.
So Y sub-2 minus Y sub-1
would be negative 3
minus negative 3
over X sub-2 minus X sub-1:
4 minus 2.
I'm going to handle that double
negative situation up there.
There we go.
So negative 3 plus 3,
that's 0.
4 minus 2, that's 2.
0 divided by 2,
that is possible to calculate.
It's 0, okay?
So if I have a line
that has a slope of 0,
then I have
a horizontal line.
And if I have
a horizontal line,
then I am concerned
with the Y's.
So if my Y's
are both negative 3,
the equation of this line
is Y equals negative 3.
And you're done.
All right,
excellent job, guys,
writing equations of lines
given two points
and still persisting and working
through those special lines,
and writing the equations
of those lines
when given two points.
Hope to see you back here soon
for more Algebra 1. Bye!