If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

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?

  • mr pink red style avatar for user Butler, Mitchell
    but how do you graph it. my algebra teacher wants me to graph it without putting it into slope intercept form.
    (27 votes)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Elsie Baum
      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.
      (20 votes)
  • leaf blue style avatar for user Kate Welch
    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?
    (44 votes)
    Default Khan Academy avatar avatar for user
  • marcimus pink style avatar for user sonia
    how would you know if the line is a parrallel line
    (8 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user CubanMissileCrisis
    At ,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)
    Default Khan Academy avatar avatar for user
  • male robot hal style avatar for user Pranabesh
    In standard form, shouldn't A in Ax+By=C always be positive?
    (8 votes)
    Default Khan Academy avatar avatar for user
  • sneak peak green style avatar for user pi expert
    what is the point of the video😑😐😶🙄
    (1 vote)
    Default Khan Academy avatar avatar for user
    • stelly blue style avatar for user Kim Seidel
      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.
      (17 votes)
  • blobby blue style avatar for user CookieMonster
    At , 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.
    (5 votes)
    Default Khan Academy avatar avatar for user
  • winston baby style avatar for user Geemobaby
    *Which is better to use and which is easier to use?*
    (4 votes)
    Default Khan Academy avatar avatar for user
    • starky tree style avatar for user C Ethan Smith
      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.
      (6 votes)
  • piceratops ultimate style avatar for user Carl
    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)
    Default Khan Academy avatar avatar for user
    • stelly blue style avatar for user Kim Seidel
      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)
  • female robot ada style avatar for user Sostar
    It's helpful that the equations lead into each other. First is point-slope, then slope-intercept, and lastly standard form.
    (4 votes)
    Default Khan Academy avatar avatar for user

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.