How do you go about writing the equation of a line? Watch our video and find out!

# Transcript: Writing the Equation of a Line

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.

## Practice Problem 1

Here’s a practice problem. Pause the video and then we’ll 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 to -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. That means this has to be the equation of the line, y= 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 that we’ll know m, we would 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 given point for the x equals y in the equation, and that would give you an equation you can solve for b. And once you solve for b, you have full slope-intercept form.

## Practice Problem

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- five-thirds, and we get a point on the line (-2, 7) 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- five-halves. And we’re also gonna plug in the coordinates of the point x equals -2 and y = 7.

We’re gonna plug in that and this will give us an equation for b. Of course- five-thirds times 2 is positive ten-thirds. And then b would be 7- ten-thirds. We find a common denominator and then we subtract and we get the y-intercept. Now, that we have full slope intercept form, 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 line is set y=0 and solve for x. So multiply by 3 and we get an x-intercept of eleven-fifths or 2.2 as a decimal. So that is a completely algebraic way to solve the problem and that’s a perfectly correct way to solve the problem.

### Another Way to Solve

That will get the answer. I also wanted to show a completely different way of thinking about this problem, what I would call a graphical solution. So we know that it goes to 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 if we move right, then it will move down. And so we could move with a run of 3, and then drop the height by -5. That would be a slope of- five-thirds. So that would mainly move the x goes up by 3, so it goes from -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 of a run, 2 over b, that has the equal the absolute value 5 over 3. Well that allows us to solve for that little b there. We get b = 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 1, 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 + 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 required in 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.

## Practice Problem

Once you have the slope, you may 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.

### Algebraic Solution

Well again we have a couple of ways to think about this. The first solution I’m gonna show is an algebraic solution. So we need to find this slope, that’s primary. We find the slope from the two known points between -3 and -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 will plug that in to 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 so I’m gonna choose the easy one. I’m gonna plug in 1, 1. And then just solve for b.

And I get b = one-fourth. So then I get three-quarters x + one-quarter, that is 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 going to plug in to find Q.

I’m just going to plug in 7 for x, Q for y. Multiply everything out and what I get is 22 over 4, of course I can simplify that that becomes eleven-halves. So that’s one way to solve the problem. Again that is a perfectly valid way to solve the problem. Incidentally I can also write eleven-halves as 5.5.

### Graphical Solution

Algebraic solutions will always get you to an answer. But here’s another way to think about it. We’re going to 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.

Now let’s think about this. We go through the point (-3,-3), 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 two has to equal three-quarters. So we set up this proportion.

We solve for h = three-halves. Three-halves, I can write that as 1.5. Now we can figure out the height of Q. So the bottom of the slope triangle of 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.

## Summary

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 in the 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 for 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.

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