Hi and welcome. This is Anthony Varela. In this tutorial, we're going to be writing linear equations given information about slope and points on a line. So we're going to look at slope-intercept form of a linear equation. We'll look at point-slope form. And we'll also talk about parallel and perpendicular lines.
So first, let's talk about slope-intercept and point-slope forms for a linear equation. Slope-intercept form is y equals mx plus b. And it's called slope-intercept form because we need to know the slope, which is the variable m, and the y-intercept, which is the variable b, in order to construct this equation in slope-intercept form.
So taking a look at a line on a graph, well, I can go ahead and determine the slope graphically looking at the rise of 2 and the run of 1. So I know rise over run is slope, so 2 divided by 1 is 2. That's the slope of this line. And I could also locate this y-intercept here. If each of these has a unit of 1, it's 1, 2, 3, 4 is the y-intercept. So I know that the equation of this line is y equals 2x plus 4. Slope is 2 and y-intercept is 4.
Now, we also have point-slope form. And this is y minus y1 equals m times x minus x1. So once again, we see this variable m for slope. So we need to know the slope of the line in order to write its equation in point-slope form, and then we also have x1 and y1. These are coordinates to a point that is on the line. So that's the other piece of information we need to know to write an equation in point-slope form.
So taking a look at this line here, once again, we can graphically determine slope from one point to another. We have to go up 1 and over 2. So the slope is 1 divided by 2 or 1/2. And then we can identify a point on the line. So this would be 3, 2. And we can use that to write the equation in point-slope form. Here is our slope m And then we have x1, which appears here, and y1, which appears here.
So we're going to get more into that later. Let's go ahead and write down what information we need to know to write the equation in certain forms. Slope and y-intercept, we use y equals mx plus b. And if we have a slope and a point, we use y minus y1 equals m times x minus x1.
Now, another thing that I'd like to point out is what if we don't know the slope at all? And we just have two points. Well, we can use any two points on a line to calculate the slope. We just take the difference in y-coordinates, so y2 minus y1, and divide that by the difference in x-coordinates, x2 minus x1. This is your rise over your run. That will give you the slope.
So let's get into an example. A line has a slope of negative 2 and passes through the point 4, negative 3. What is the equation of this line? Well, let's first think about what information we can draw from our problem. We know that the slope is negative 2. So m equals negative 2. We're also given a point. So I know that my x-coordinate is 4, and my y-coordinate is negative 3.
So because I know the slope and a point, I'm going to use point-slope form to develop my equation. So we can go ahead and substitute then negative 2 is the slope. So we have that here. And then we need to put in our x1 and y1. So y1 goes over here, negative 3. And x1 goes right here, 4.
So we have our equation written in point-slope form. But oftentimes, we prefer our equation to be written in slope-intercept form, y equals mx plus b. So we have y on one side of the equation. So let's go ahead and rewrite this.
Well, the first thing I want to do is rewrite y minus negative 3 as y plus 3. Subtracting a negative is the same as adding a positive. Next, what I want to do is distribute that negative 2 into x and minus 4. So I have y plus 3 equals negative 2x plus 8. Now, I can subtract 3 from both sides to get y equals negative 2x plus 5. So now, this is the same line just written in a different form, slope-intercept form.
And we can even go ahead and double check that we've done this right. So here's our equation in slope-intercept form. And we can use the point that we know for 3 to plug in 4 for x-- or for negative 3, excuse me, 4 for x, and negative 3 for y, and see if this is a true statement. So evaluating this, we have negative 2 times 4 is negative 8. And when we add 5 to that, we get negative 3. And negative 3 equals negative 3. So we've done this right.
Now, our next example-- this has given two points. So a line passes through the points negative 2, 5, and 3, 15. What is the equation of the line? Now, notice we don't know at all right now what the slope of this line is. It hasn't been given to us.
So we need to calculate the slope. So we're going to be using our formula for slope. We need an x1 and y1. That's going to be negative 2 and 5. And we need x2 and y2. That's going to be 3 and 15. And we're going to plug that into our equation for slope.
So plugging in our y values, this gives us 15 minus 5. And then plugging in our x values, that gives us 3 minus negative 2. Well, this then simplifies to 10 over 5. 15 minus 5 is 10. And 3 minus negative 2 is 3 plus 2, which is 5. So our slope is a positive 2.
So now, we know the slope. And we also then need to use one of our points. And it doesn't matter which one we use. So let's just choose one. And we'll substitute our x and our y. So I've chosen to use this point negative 2, 5. So here is our negative 2. Notice this is a plus 2 in our equation because our general point-slope form is x minus x1. So be careful about that. And then I'm plugging in a 5 here.
So now, we have our equation written in point-slope form. Let's once again go ahead and convert this into slope-intercept form. So I'm going to distribute that 2. So I have y minus 5 equals 2x plus 4. And then I'll add 5 to both sides. So I have y equals 2x plus 9. We can go ahead and use our other point to check and make sure that this is correct.
So I'm plugging in 15 for y and plugging in 3 for x. Now, does 15 equal 2 times 3 plus 9? Well, 2 times 3 is 6. And when we add 9 to that, we get 15. So we have a true statement here. We've done this right.
Our last example says that a line is parallel to y equals 5x minus 4 and passes through the point 2, 13. What is the equation of this line? Well, we need to know the relationship between parallel lines and perpendicular lines if this situation was about a line perpendicular to y equals 5x minus 4.
Well, we know our point, but what about the slope? Well, parallel lines have identical slopes. And perpendicular lines have opposite reciprocal slopes. So here, we're dealing with a line parallel to y equals 5x minus 4. So the slope, we know, is 5 because m stands for slope. And so a line parallel would have the same slope. So we know that the slope is also 5.
So we have our slope. And we have a point. Let's go ahead and use our slope and point to write an equation in point-slope form. So we have y minus 13 equals 5, the slope, times x minus 2. That's their equation in point-slope form.
And we'll go ahead and rearrange this so we have it in slope-intercept form by distributing that 5. y minus 13 equals 5x minus 10. And adding 13 to both sides, we have y equals 5x plus 3. Checking to make sure that we've done this right, I can put in 13 for y and 2 for x. 5 times 2 is 10. And when we add 3, we get 13. So we've done this right.
So let's review writing a linear equation using slope and points. We talked about if you know the slope and the y-intercept, you can write it as y equals mx plus b. If you know a slope and a point, you can write it as y minus y1 equals m times x minus x1. And if you don't know the slope at all, but you know two points, you can calculate the slope using the difference in y over the difference in x.
And remember, parallel lines have identical slopes. And perpendicular lines have opposite reciprocal slopes. Thanks for watching this tutorial on writing a linear equation using slope and points. Hope to see you next time.