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## Algebra 1

### Course: Algebra 1 > Unit 5

Lesson 6: Summary: Forms of two-variable linear equations# Writing linear equations in all forms

Sal finds the equation of a line that passes through (-3,6) and (6,0) in point-slope, slope-intercept, and standard form. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- but how do you graph it. my algebra teacher wants me to graph it without putting it into slope intercept form.(23 votes)
- Well, say the equation is 8x -2y =24. To graph, you must plug in 0 for either x or y to get the y- or x-intercept. So in the equation that I said, let's find the y-intercept first. You would plug in 0 for x. So the equation would be 8*0 -2y =24, or -2y =24. Then you can solve it like a regular equation and you would get y =-12. For the x-intercept, it's basically the same thing, except you plug in 0 for y instead of x. So you would get 8x -2*0 =24 or 8x =24. Once again, you would solve it like a regular equation, and get x =3. So the y-intercept is -12 and the x-intercept is 3. Then you can use those two points [(3,0) and (0,-12)] to find the slope and graph from there. I know this is a little late and you've probably figured it out by now, but I'm still posting this for those out there who had the same question and have not figured it out.(12 votes)

- He says 'if you WANT to make it look extra clean' to get rid of the fraction, but isn't one of the rules of Standard Form that you can't have fractions? Wouldn't you have to get rid of that fraction anyway?(38 votes)
- You wouldnt have to. It would really just depend on how your professor would like the form to be.(8 votes)

- how would you know if the line is a parrallel line

y=2/3x-2(8 votes)- well if slope of line 1 is equal to slope of line 2 they are parallel

lets say if equation of line 1 is y=m1x+c

and line 2 is y=m2x+c

then m1 and m2 should be equal in order to make them parallel

m1=m2(8 votes)

- At7:25,Sal says that the equation is in standard form.I thought you couldn't have fractions in standard form.Can someone explain please?(13 votes)
- I'm not sure, but the way I learned it, you are right. it should be 2x+3y=12(4 votes)

- In
**standard form**, shouldn't**A**in*Ax+By=C*always be**positive**?(8 votes)- Not necessarily.

A Linear equation in standard form is written as**Ax + By = C**, This*does not*mean that**A**should always be Positive. But by convention, the equation is written in a way that we get**A >= 0**.

You can find more info at Wikipedia (http://en.wikipedia.org/wiki/Linear_equation) or by simply running a Google search.(11 votes)

- At4:15, Sal says he will use the point (-3,6) for the point slope form. Does it matter what point you use for the point slope form? In some of the Khan Academy exercises, the questions say I am wrong when I use different points for point-slope form.(6 votes)
- It does not matter which point you use in point slope form. If your answer is marked wrong, then you much have an error. If you tried posting your problem and your work, then someone could help you find your error.(4 votes)

- what is the point of the video😑😐😶🙄(1 vote)
- Despite the fact that you just called me a "nerd" in a comment you posted on another entry, I'm going to try and answer you question anyway. Maybe it's time for you to find some of your own "inner nerd".

What is the point of the video? Linear equations can appear in different forms. Each form has it's own benefits. This video is showing you the different forms and how you can create the equation for a line given two points that are on the line.(9 votes)

- *Which is better to use and which is easier to use?*(3 votes)
- I think y=mx+b is the easiest formula. I think it is the easiest because you can easily graph it, also if you need to change it into the other formulas it can be done easily.
*But everyone has different opinions so find the best that works for you*, good question.(5 votes)

- At 4.33, Sal uses
**6**as his**b**for the point slope mode: y -**b**= mx (x-a) -> y -**6**= -2/3(x--3).

But at 5.49 he uses mx * a to define his b for the slope intercept mode. And therefore his**b**ends up being**4**in the final slope intercept mode: y = mx +**b**-> y = -2/3x+**4**.

When y= mx+**b**, why is y = -2/3 +**6**not a valid answer? This was my natural instinct, when i tried to solve for the slope intercept mode before the point slope mode.(2 votes)- In the point slope form, Sal uses "b" as a regular variable to represent the y-value in an ordered pair of the form (a, b). He is not using "b" at this time as the y-intercept. Remember, a y-intercept will always have an X-value = 0 because the point must sit on the y-axis. The point (-3, 6) that Sal used to find the equation clearly is not on the y-axis, so it can not be the y-intercept for the line.

Once the equation is changed into slope-intercept form, the y-intercept has been calculated as (0, 4).

Hope this helps.(6 votes)

- How do you turn a linear equation like y=-2+1/4 into a standard form?(3 votes)
- The format for standard for is y-mx=b. so for this specific equation it would be y+2x=1/4.(2 votes)

## Video transcript

A line passes through the points
negative 3, 6 and 6, 0. Find the equation of this line
in point slope form, slope intercept form, standard form. And the way to think about
these, these are just three different ways of writing
the same equation. So if you give me one of them,
we can manipulate it to get any of the other ones. But just so you know what these
are, point slope form, let's say the point x1, y1 are,
let's say that that is a point on the line. And when someone puts this
little subscript here, so if they just write an x, that means
we're talking about a variable that can take
on any value. If someone writes x with a
subscript 1 and a y with a subscript 1, that's like saying
a particular value x and a particular value of y,
or a particular coordinate. And you'll see that when
we do the example. But point slope form says
that, look, if I know a particular point, and if I know
the slope of the line, then putting that line in point
slope form would be y minus y1 is equal to
m times x minus x1. So, for example, and we'll do
that in this video, if the point negative 3 comma 6 is on
the line, then we'd say y minus 6 is equal to m times x
minus negative 3, so it'll end up becoming x plus 3. So this is a particular
x, and a particular y. It could be a negative
3 and 6. So that's point slope form. Slope intercept form is y is
equal to mx plus b, where once again m is the slope, b is the
y-intercept-- where does the line intersect the y-axis--
what value does y take on when x is 0? And then standard form is the
form ax plus by is equal to c, where these are just two
numbers, essentially. They really don't have
any interpretation directly on the graph. So let's do this, let's figure
out all of these forms. So the first thing we want to do
is figure out the slope. Once we figure out the slope,
then point slope form is actually very, very, very
straightforward to calculate. So, just to remind ourselves,
slope, which is equal to m, which is going to be equal to
the change in y over the change in x. Now what is the change in y? If we view this as our end
point, if we imagine that we are going from here to
that point, what is the change in y? Well, we have our end point,
which is 0, y ends up at the 0, and y was at 6. So, our finishing y point is 0,
our starting y point is 6. What was our finishing x
point, or x-coordinate? Our finishing x-coordinate
was 6. Let me make this very clear, I
don't want to confuse you. So this 0, we have that 0, that
is that 0 right there. And then we have this 6, which
was our starting y point, that is that 6 right there. And then we want our finishing x
value-- that is that 6 right there, or that 6 right there--
and we want to subtract from that our starting x value. Well, our starting x value is
that right over there, that's that negative 3. And just to make sure we know
what we're doing, this negative 3 is that negative
3, right there. I'm just saying, if we go from
that point to that point, our y went down by 6, right? We went from 6 to 0. Our y went down by 6. So we get 0 minus
6 is negative 6. That makes sense. Y went down by 6. And, if we went from that point
to that point, what happened to x? We went from negative 3 to
6, it should go up by 9. And if you calculate this, take
your 6 minus negative 3, that's the same thing as
6 plus 3, that is 9. And what is negative 6/9? Well, if you simplify it,
it is negative 2/3. You divide the numerator and
the denominator by 3. So that is our slope,
negative 2/3. So we're pretty much ready
to use point slope form. We have a point, we could pick
one of these points, I'll just go with the negative 3, 6. And we have our slope. So let's put it in
point slope form. All we have to do is we say y
minus-- now we could have taken either of these points,
I'll take this one-- so y minus the y value over here, so
y minus 6 is equal to our slope, which is negative
2/3 times x minus our x-coordinate. Well, our x-coordinate, so x
minus our x-coordinate is negative 3, x minus negative
3, and we're done. We can simplify it
a little bit. This becomes y minus 6 is equal
to negative 2/3 times x. x minus negative 3 is the
same thing as x plus 3. This is our point slope form. Now, we can literally just
algebraically manipulate this guy right here to put it into
our slope intercept form. Let's do that. So let's do slope intercept
in orange. So we have slope intercept. So what can we do here
to simplify this? Well, we can multiply out the
negative 2/3, so you get y minus 6 is equal to-- I'm just
distributing the negative 2/3-- so negative 2/3 times
x is negative 2/3 x. And then negative 2/3 times
3 is negative 2. And now to get it in slope
intercept form, we just have to add the 6 to both sides so
we get rid of it on the left-hand side, so let's
add 6 to both sides of this equation. Left-hand side of the equation,
we're just left with a y, these guys cancel out. You get a y is equal
to negative 2/3 x. Negative 2 plus 6 is plus 4. So there you have it, that is
our slope intercept form, mx plus b, that's our
y-intercept. Now the last thing we need
to do is get it into the standard form. So once again, we just have to
algebraically manipulate it so that the x's and the
y's are both on this side of the equation. So let's just add 2/3 x to both
sides of this equation. So I'll start it here. So we have y is equal to
negative 2/3 x plus 4, that's slope intercept form. Let's added 2/3 x, so
plus 2/3 x to both sides of this equation. I'm doing that so it I don't
have this 2/3 x on the right-hand side, this
negative 2/3 x. So the left-hand side of the
equation-- I scrunched it up a little bit, maybe more than I
should have-- the left-hand side of this equation is what? It is 2/3 x, because 2 over
3x, plus this y, that's my left-hand side, is equal to--
these guys cancel out-- is equal to 4. So this, by itself, we are in
standard form, this is the standard form of the equation. If we want it to look, make it
look extra clean and have no fractions here, we could
multiply both sides of this equation by 3. If we do that, what do we get? 2/3 x times 3 is just 2x. y times 3 is 3y. And then 4 times 3 is 12. These are the same equations,
I just multiplied every term by 3. If you do it to the left-hand
side, you can do to the right-hand side-- or you have to
do to the right-hand side-- and we are in standard form.