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Writing Equations of Lines

Writing Equations of Lines. Sometimes, the test will give you information and ask you to come up with the equation of a line. In other questions, 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’ll talk about this. Okay, the line shown passes through the point A comma 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, tend to goes to negative 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 plus 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 equals 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 equals mx plus b, so we would 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 to the given point for the x equals 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-sevenths and we want to 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 equals mx plus 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 two and y equals negative seven.

We’re gonna plug that in, this will give us an equation for b. Of course negative five-thirds times negative two is positive ten-thirds. And then b would be seven minus ten-thirds. We find a common denominator and then we subtract and we get the, the y-intercepts. So now we have the full slope intercept form. The whole y equals mx plus b form.

Now what we need to find the x-intercept. Well, of course the way we find the x-intercept of a line is to set y equal to 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.

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 to the point negative 2 comma 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 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 comma 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 a run. 2 over b, that has to equal the absolute value 5 over 3. Well that allows us to solve for that, 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 1. And then the distance that I’m pulling b there is an ori, is a distance of six-fifths. So, we’ll just add, so 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, 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 coordinate 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 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 negative 3 comma negative 2, 1 comma 1 and 7 comma Q. Find the value of Q. Well again, we have a couple ways of thinking about this. This 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 negative 3, negative 2, and 1 comma 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 in to y equals mx plus b, we can plug either one of the points in.

I’m going to say that plugging 1 1 is gonna be much easier than plugging in negative 3 comma negative 2. I get to choose, so I’m going to choose the easier one. 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 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, 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 could 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 you can ever, you, 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 negative 3 comma 2, we go through the point 1 comma 1. We wanna get a little closer to that point 7 comma Q, so I’m gonna go over 4 and up 3 and so that will put me at 5 comma 4, and then I’m reasonably close. So now from 5 comma 4, of course the, 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 of a 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 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 in this diagram is at a height of 4. And then to get up to Q it’s just 4 plus h. Or 4 plus 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 the slope. Once we have a point in the slope, we can plug these into y equals mx plus 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.