 Here's an example of writing equations and lines in the three different forms. We want to write the equation of a line that passes through the point negative 6 5 and has a slope of one half. And we want to do so in the three separate forms, point, slope, form, slope, intercept, and standard form. First remind yourself, make sure you remember what all those forms look like. So we'll have three different equations and they all describe the same line. So let's start with point, slope, form. In order to write the equation of a line in point, slope, form, we need a point, which we have, and a slope, which we also have. So let's rewrite those. And now all that's left to do is to substitute these numbers into into the point, slope, form. So we have y minus y1 and remember y1 was 5. So y minus 5 is equal to slope, which was one half, times the quantity x minus x1. And x1, of course, is negative 6. So let's write the whole thing in negative 6. But we can simplify x minus negative 6 one step. When you subtract a negative, that's equivalent to adding. So we can rewrite this as x plus 6. And there we go. Point, slope, form for this line. Now let's move on to slope intercept form. First, let's write what slope intercept form looks like in general. We need an equation in the form y equals mx plus b. In order to get there, let's use what we've done previously using point, slope, form. So using a bit of algebra, we can rewrite this point, slope, form into slope intercept form. So in order to get mx plus b, we'll have to distribute the one half. And in order to get y alone, we'll have to add 5. So let's start by using the distributive property. So we have one half times x and one half times 6. And then let's add 5 to both sides of the equation. And there we go. We have an equation in the form y equals mx plus b. We see that m is one half and the y intercept would be 3. Now let's move on to standard form. First, remind yourself what standard form is. We have a times x plus b times y equals c. a, b, and c must all be integers. And in particular, the coefficient with x must be positive. So we'll use the slope intercept form in order to, again, using algebra to get that equation into standard form. So let's rewrite the slope intercept form. So first thing we do, well, we want to get the x's and y's together on one side of the equation. So let's subtract one half x. And now remember in standard form, each of the coefficients, and by coefficients I mean these a, b, and c numbers, each of them must be integers. Well, right now we have a one half. And so if we multiply both sides of the equation by 2, that one half will be, well, 2 times half is equal to 1. So let's do that. We'll multiply both sides by 2. So we have 2 times negative one half x is negative x or negative 1x. We'll just say negative x. 2 times y, so we'll add 2y. And that equals 2 times 8, which is 16. Now, lastly, in standard form, the coefficient with x is supposed to be a positive number. And right now it's a negative, so negative 1. So in order to make a negative number positive, we can multiply both sides of the equation times negative 1. So negative 1 times negative x, that's just positive x. Negative 1 times 2y, that's negative 2y. And negative 1 times 16, of course, is negative 16. Now, you might be thinking to yourself, wait a minute, is it OK to have negative numbers? Negative 2, negative 16, that's acceptable in standard form. A, right, the coefficient with x is supposed to be positive. B can be negative and C can be negative as well. So here we have a plus negative, sorry, x plus negative 2 times y is equal to negative 16. And now we're done.