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Skip to 0 minutes and 13 seconds In this activity, we’ll be looking at some interesting problems involving lines and linear relationships. So we’re going to be looking at a problem going back to the ancient Chinese, which is geometrically how to find the intersection of two lines. They also looked at more general systems where you have more than one linear equation. We’ll talk about conversion from one system of units to another, a very important practical kind of thing to be able to do. And we’ll look at some finance, how interest rates and principles relate, and give you some problems to work on. So there’s some nice challenges here that will strengthen your algebraic skills as well as your geometrical intuition.

Skip to 0 minutes and 59 seconds The first problem that we want to have a look at is a very fundamental one, which is how to find the equation of a line through two given points. So we’re asked to find the line through the point A, which is [3,2], 3 this way, 2 that way. And the point B, which is minus 4 minus 1. That’s minus 4 in the x direction and minus 1 in the y direction down here. So there geometrically is the line. It’s easy enough to draw once we draw the two points, but what is the equation of that line? Well, we’re going to look for an equation for the line of the form y equals mx plus b.

Skip to 1 minute and 44 seconds That’s our fundamental form for the line. So m, I remind you, represents the slope of that line. And b represents the y-intercept where the line crosses the y-axis. So let’s find m first, the slope, which is really the most fundamental constant associated to the line. How do we find the slope? Well, the slope, which is m, sometimes it’s written as delta y over delta x representing the ratio between the change in the y-coordinates over the change in the x-coordinates. So we go from here to here say, then this change in x down here is– let’s see, we’re going from minus 4 to 3. So it’ll be 3 minus, minus 4. That’s a total of 7.

Skip to 2 minutes and 40 seconds While the change in y, as we go from here to here, we’re going from minus 1 to 2. So again, it’s the difference, 2 minus, minus 1. So that’s a total of 3. So the change in y over the change in x, that’s 3 over 7. That’s the slope of that line. You can see that’s geometrically quite reasonable. It’s less than a slope of 1. Slope of one would be like 45 degrees.

Skip to 3 minutes and 10 seconds So now we know that the equation is y equals 3/7 of x plus some unknown b. That’s still to be determined. So how are we going to find b, this y-intercept? What we’re going to do is we’re going to substitute one of these points into here, because we know that both of these points must satisfy this equation.

Skip to 3 minutes and 38 seconds So sub in [3,2] to get– we replace x with 3 and y with 2, we get two equals 3/7 times 3 plus b, or b equals 2 minus 9 sevenths, which is 14 minus 9 is 5/7.

Skip to 4 minutes and 13 seconds So we finish that y equals 3/7 x plus 5/7, and that’s the equation of the line, is the equation of our line l. And if we wanted to check it, we could substitute the other point in and just make sure that you replace x with minus 4 and y with minus 1, that the equation’s actually satisfied. So a very basic fundamental technique in this theory to be able to write down efficiently the equation of a line given two points on it. First, we find the slope as a change in y over change in x, then we can substitute a point to find the y-intercept.

Skip to 5 minutes and 4 seconds Our next problem is one with a great history, the problem of finding the intersection of two lines or algebraically solving a pair of linear equations. So, in particular, with this question, we’re asked to find the intersection of the lines l, this line here, with the equation y equals 3/7 x plus 5/7– that’s just the equation that we found on the last slide– and this new line in green. Its line, its name is m and has equation y equals minus x plus 3.

Skip to 5 minutes and 35 seconds Now let’s note that this represents a line of slope minus 1, which means that if you go over one unit in the x direction, the y value goes down correspondingly 1 and that this three represents this y-intercept 3 that’s pictured there. So geometrically we could draw these two lines, and we could guess roughly what this point of intersection is. We want to find this point algebraically. So we need to solve this pair of equations, y equals 3/7 x plus 5/7 together with y equals minus x plus 3. We want to find a point xy, which is really this point right here, this unknown point xy, which satisfies both equations.

Skip to 6 minutes and 32 seconds So how do we do that? Well, we eliminate the y from this pair of equations, because we have y equals this and y equals this. So if both equations are satisfied, then it must be true that 3/7 x plus 5/7 equals minus x plus 3. The y value here equals the y value there, and this is then only an equation involving x. So we can relatively easily solve that. How do we do that? Well, let’s get the x’s to one side. If we bring that x to the other side here, then we will get all together 10/7 of x. And if we bring the 5/7 to the other side, that’s 3 minus 5/7, which is 21/7 minus 5. That’s 16/7.

Skip to 7 minutes and 30 seconds So therefore, x equals– 7’s, you multiply through and we get half x equals 16 over 10, which is the same as 8 over five.

Skip to 7 minutes and 48 seconds So there is the x value. What about the y value? Well, we have to then take that x value and plug it back into either one of these equations to find the corresponding y value. So then y equals minus 8/5 plus 3, that equation there. And that is then the same as 3 minus 8/5, which is 15 minus 8. That’s 7/5.

Skip to 8 minutes and 25 seconds There’s the y value.

Skip to 8 minutes and 29 seconds So the point of intersection is the point x-coordinate’s 8/5 and y-coordinate’s 7/5. And there it is right there. So 8/5 somewhere between 1 and 2. 7/5 also between 1 and 2 there.

Solving problems with lines and linear relations

We show how to find the equation of the line through two given points, and how to algebraically solve for the intersection of a pair of lines.

Finding the line between two points

Finding the equation of the line \(\normalsize{y=mx+b}\) through two given points proceeds in two steps: we find the slope \(\normalsize{m}\) and then we find the \(\normalsize{y}\)-intercept \(\normalsize{b}\).

To find the slope \(\normalsize{m}\) of the line, we just take the change in \(\normalsize{y}\) divided by the change in \(\normalsize{x}\). Once we have \(\normalsize{m}\), we substitute one of the points into the equation to find \(\normalsize{b}\).

Finding the point on two lines

Solving a pair of linear equations to find their meet, or intersection, goes back to the ancient Chinese. The video shows how to eliminate one of the variables (often \(\normalsize{y}\)) to get a single linear equation in the single variable \(\normalsize{x}\). Solving that to find \(\normalsize{x}\) is then easy, and then we substitute back into one of the original equations to find \(\normalsize{y}\).

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Maths for Humans: Linear, Quadratic & Inverse Relations

UNSW Sydney