 Our task in this video is to write the equation of a line in point-slope form, in slope-intercept form, and in standard form for this graph. So we have a line, and we need to write the equation based on that line. So first off we'll start using the point-slope form. Now point-slope form, again as the name suggests, we need a point and we need a slope. First, remind yourself what point-slope form is by writing out what the equation of that general form is. So in order to use the point-slope form we'll need a point and we'll need the slope. So let's just pick one of these two points that are shown here. It doesn't matter which if we use this one, let's say that'll be just fine. So that's the point one, two. Next we need the slope of that line. In order to find the slope we'll need the slope formula, which is a formula that you should write down if it's not in your notes. It should be. So let's figure out what that formula is. So we have y2-y1 divided by x2-x1. Remember y's are on the top and x's are on the bottom. So we'll use these two points and we'll plug in their coordinates. So we've got 3-2 and then we'll divide by negative 2-1. So that gives us 1 divided by negative 3 and there's our slope. And all that's left to do is to substitute the slope and the point into point-slope form. So once again we have y-y1, which we know is 2. So we'll fill that in. y-2 is equal to the slope, which we found to be negative 1-third times the quantity x-x1. x1 was 1, so we'll just plug in a 1. So we have y-2 is equal to negative 1-third times the quantity x-1. So now we'll use the point-slope form for the slope-intercept form. So slope-intercept form, once again let's remind ourselves what that form looks like. And just like in the previous video, we'll use the point-slope form to find the slope-intercept form. So recall what the point-slope form was from the previous problem. So using algebra, we'll take the point-slope form and we'll turn it into something that looks like mx plus b and we'll isolate y as well. So to start off let's use the distributive property on negative 1-third times the quantity x-1. So negative 1-third times x of course is just negative 1-third x. And then negative 1-third x times negative 1, we're going to add plus 1-third. And that still equals y-2. And now we'll add 2 to both sides in order to isolate y by itself. So we're left with y equals negative 1-third x plus 2 and 1-third. And so now we have our equation in slope-intercept form. Next let's take that and convert it into standard form. So we want to take our slope-intercept form and turn it into standard form. Remember a, b, and c are all integers and in particular a is a positive integer. So as it stands right now we've got y on one side of the equation and x on the other side. Let's fix that and add the opposite, add 1-third x to both sides of the equation. And now remember these coefficients have to be integers. The coefficients are the numbers a, b, and c. Here I've got 1-third, here I have 2 and 1-third and neither of those are integers. So I'll need to multiply both sides of the equation by the denominator. Here I've got 3, I've also got 3. So if I multiply both sides of the equation by 3 then those fractions will become integers. We'll distribute the 3. So 3 times a third is just 1. So we're left with x. 3 times y is 3y and that equals 2 and 1-third times 3. So 2 and 1-third times 3. 2 and 1-third is the same as 7-thirds. And 7-thirds times 3 is 21-thirds. And 21 is divisible by 3 which equals 7. So that tells me that this side of the equation, 2 and 1-third times 3 is 7. So standard form of that line. So a quick recap in point-slope form. The equation of that line is y-2 is negative 1-third times the quantity x-1. Slope intercept form is y equals negative 1-third x plus 2 and 1-third. And standard form is x plus 3y is equal to 7.