## Writing Equations of Lines

### Transcript

Writing equations of lines. Sometimes, the test will give you information and ask you to come up with the equation of a line. Another question, solving for the equation of the line will be very helpful in finding what the question asks. The test could expect you to find the equation of a line either from numerical information, here's the value of the slope, here's a point, here's the y intercept, that sort of thing.

Or they could give you a picture. If you are given a picture of the line, it should be relatively easy to read both the slope and the y-intercept from the picture, and these will allow you to create the equation. Here's a practice problem. Pause the video, and then we will talk about this.

Okay, the line shown passes through the point A, 30. What is the value of A? Well, it's very easy to read off this graph, the value of the y-intercept. The y-intercept is 1. And the slope, well, since it goes through -2 we have a run of 2 and a rise of 1, so rise over run would be one-half.

So the slope is one-half and the y-intercept is 1. So that means this has to be the equation of the line y equals one-half x+1, well, now we have an equation. So now we can plug in. We plug in A for x and 30 for y. Subtract 1, and then multiply by 2 to cancel the one-half.

And we get A=58, and that's the answer. The test could give you the slope and then some point on the line. The slope and a single point are enough to determine a unique line. You can plug the slope directly into the slope intercept form, y=mx+b. So we would know m, initially not know the value of b but, because every point on the line must satisfy the equation of the line, you can plug in the coordinates of the give point for the x=y in the equation, and that will give you an equation you can solve for b.

And once you solve for b, you have full slope intercept form. So here's a practice problem. Pause the video, and then we'll talk about this. Okay, so what we're given here is a slope. We have a slope of negative five-thirds, and we get a point on the line negative two-seventh, and we wanna know what is the x-intercept of the line.

Well, we have two very different ways of going about this. And I'm gonna show them both. The first is what I would call an algebraic solution. So here, we're gonna write y=mx+b. We know the m. So, we're gonna plug in the slope of negative five-halves and we're also gonna plug in the coordinates of the point, x equals negative 2 and y equals 7, we're gonna plug that in.

This will give us an equation for b. Of course, negative five-thirds times negative 2 is positive ten-thirds, and then b would be 7 minus ten-thirds. We find a common denominator and then we subtract and we get the y-intercept. So now we have the full slope intercept form, the full y=mx+b form. Now what we need to find is the x-intercept.

Well, of course, the way we find the x-intercept of a line is to set y equal to zero, and solve for x. So multiply by 3, and we get an x-intercept of eleventh-fifths, or 2.2, was a decimal. So that is a completely algebraic way to solve the problem, and that's a perfectly correct way to solve the problem.

That will get the answer. I also wanna show a completely different way of thinking about this problem, what I would call a graphical solution. So we know that it goes through the point -2,7. Well, let's think about this slope, what does that slope mean? If we wanna get closer to the y-intercept, because it has a negative, the x-intercept, we need to move to the right, because it has a negative slope.

So we move right, then it will move down. And so we could move with a run of 3 and then drop the height by negative 5. That would be a slope of negative five-thirds. And so, that would mean we move the x goes up by 3. So it goes from negative 2 to positive 1. And the y goes down by 5.

So it goes from 7 down to 2. So now we're a lot closer to the x-axis. Now let's think about that visually. We have the point 1,2, which is on the line. Then there's the x-intercept. And notice that little triangle that we make there, that has to be a slope triangle.

So in other words, if we just look at the absolute values, 2 over b, that's a rise over run, 2 over b. That has to equal the absolute value 5 over 3. Well, that allows us to solve for that little b there. We get b equals six-fifths. Now, let's think about this.

We want the x-intercept. Well, we know the distance from the origin to 1,0, that's a distance of one. And then the distance that I'm pulling b there is a distance of six-fifths. So we'll just add to the six-fifths. I'll write that as 1.2 for simplicity. 1.2 plus 1, that would be an x-intercept of 2.2.

So that is a graphical way of approaching this. Now that might be an unfamiliar way. That's typically not the way they teach you to approach things in school, but I would point out, if you can solve the problem algebraically and also solve it graphically using proportions, then you really understand it very deeply. It will enormously enhance your understanding of ordinate geometry to think about both approaches.

The test could give you two points and expect you to find the equation of the line. Two points uniquely determine a single line. So if you're given the two points, what do you do? Well, of course, from the two points, we can always find a slope. And then once we have the slope again what we'll do, we'll plug either one of those points, it doesn't matter, either one of those points for the x and the y in the equation, plug the slope in, we can solve for b, and then we have full slope intercept form.

Once you have the slope you may be able to think about the question graphically as well. Here's a problem. Pause the video, and then we'll talk about this. Okay, line J passes through the points -3,-2, 1,1 and 7,Q.

Find the value of Q. Well again, we have a couple of ways of thinking about this. The first solution I'm gonna show is an algebraic solution. So we need to find the slope that's primary. We find the slope from those two known points between -3,-2 and 1,1. We have a rise of 3 and a run of 4.

So rise over run is three-fourths. So that's the slope. Now, we'll plug that into y=mx+b. We can plug either one of the points in. I'm gonna say that plugging 1,1 is gonna be much easier than plugging in -3, -2. I get to choose, I'm gonna choose the easier one.

I'm gonna plug in 1,1 and then just solve for b. And I get b equals one-quarter. So then I get three-quarters x plus one-quarter that is the slope intercept form. So now I have the equation of the line itself. I have the equation of line J.

Well now, I'm just gonna plug in to find Q. I'm just gonna plug in 7 for x, Q for y, multiply everything out, and what I get is 22 over 4, which I can simplify that, that becomes eleven-halves. So that's one way to solve the problem. Again, that's a perfectly valid way to solve the problem.

Incidentally, I can also write eleven-halves as 5.5. Algebraic solution will always get you to an answer. But here's another way to think about it. We're gonna use a graphical solution. So again, you have to find the slope. The slope is primary information about the line.

If it's available to find the slope, always find the slope, it always helps you. All right, so now let's think about this. We go through the point -3, -2, we go through the point 1,1. We wanna get a little closer to that point 7, Q. So I'm gonna go over 4 and up 3 and so that will put me at 5,4 and then I'm reasonably close.

So now from 5,4, of course, the run going from 5 to 7 is 2. So that little slip triangle there has a run of 2, we'll call the height of it h and it must be true that rise over run, h over 2 equals the slope of the line. H over 2 has to equal three-quarters.

So we set up this proportion. We solve for h. H equals three-halves. Three-halves, I can write that as 1.5. So now we can figure out the height of Q. So the bottom of the slope triangle this diagram is at a height of 4.

And then to get up to Q, it's just 4+h or 4+1.5, which gives us 5.5 so that is a graphical way of solving the exact same question. Again, when you're doing your practice, if you can solve it algebraically and also use proportions to solve it graphically you will have a much deeper understanding of coordinate geometry.

If we are given a well-labeled graph, we may be able to read the slope and the y-intercept from the graph itself. That's a big idea. Once we are given two points, we can find a slope. Once we have a point and a slope, we can plug these into y=mx+b to solve for b. And remember, you often have the option of solving algebraically or graphically, thinking about the proportions involved with the slope.

And again, if you can solve the problem both ways, with an algebraic solution and a graphic solution, you understand this topic extremely well.