We already know a lot about lines in the plane. Let’s now complete our understanding of this topic. Let me recall that a function f is called affine, if its defining formula is of the form mx plus b for certain parameters m and b. As we know from week 2, the graph of an affine function is a line. m is called the slope of the line, the parameter m, and if m is strictly positive, you get an increasing function. Negative gives a decreasing function. And if m is 0, you have a constant function, the line is horizontal. As for b, the y-intercept, it simply is the y-coordinate of the point at which the line intersects the y-axis.
So to summarize, in part, the graph of an affine function is a non-vertical straight line in the plane. It is a fact that the converse of this result is true. Any non-vertical straight line in the plane is the graph of a certain affine function, f. Now given the line which corresponds clearly to some function f because it’s not vertical, how do we prove that f is of the form mx plus b? And how do we find m and how do we find b? Well as far as finding b is concerned, it’s rather simple. Because when you plug x = 0 into the expression mx plus b, you get b.
In other words b has no choice but to be the value of the function at 0. However m is a little more subtle. Let’s see how to find it. I take the straight line we have, the graph of this function f, and I take any two points on the line, two distinct points. These two points will generate a horizontal increment, the difference in the x-coordinates that I call delta x. And they will generate also a vertical increment that I call delta y. Notice that delta x is not 0 because the points are distinct. The delta y could be 0 if the line is horizontal, that would be fine.
I claim now that the slope, the m, is generated by dividing delta y by delta x. I claim, to be more precise, that whatever two distinct points you take on the line will lead to the same value of m. Let’s see why this is the case. We’ve calculated m with the two indicated points. Suppose now I take any other two distinct points on the line. They will generate of course different horizontal and vertical increments. I use the capital delta here to indicate them. Capital delta y and capital delta x. However, you notice that the two triangles in the picture are similar. Similar in the sense of elementary geometry. That implies that the proportions of their relative sides are maintained.
To be more specific, capital delta y over a capital delta x is the same as little delta y over little delta x. That is, you get the same value of m. Now to exploit this observation, let’s take our two points to be the points (0, f(0)) and (x, f(x)) where x is any non-zero point. Then, if you write down the fact that the vertical increment is f(x) minus f(0), clear, and the horizontal increment is x minus 0 or x. And if you write them that this ratio gives you m, multiply across by x, you find f(x) is equal mx plus f(0). This is true, of course, for x equals 0 as well.
And we have proved then that f is an affine function and that its slope m can be calculated as we have pointed out. Let’s observe a consequence of our discussion so far. Suppose I have two fixed parameters, c and d, not both 0. And I consider the set of points, (x,y) and the plane, whose coordinates x,y satisfy the linear equation cx plus dy equals q, q being another given fixed parameter. I claim that the set of those points is precisely a straight line. In other words, whenever you have this linear relationship between the coordinates, then the set of solutions is the set of points on a straight line. We can see this a little more clearly.
If d is different from 0, we can actually solve the equation for y. And we see that what we have is the graph of an affine function, so a non-vertical straight line. In fact, the slope of the straight line will be minus c over d. On the other hand, if d is equal 0, then c is different from 0. And you can solve for x and you get x equals q over c. In other words, we’re looking at all points with a prescribed x-coordinate and any y-value whatever. That describes a vertical line in the plane. Now we can exploit these facts that we’ve just discussed to obtain a geometric understanding of a previous result that we had in week four.
Now a geometric understanding, to my mind, is something that’s deeper, more profound, than a merely linguistic type of understanding. Here’s the result that I’m referring to. You may recall that we had studied a system of two linear equations in two variables, x and y. We had referred to the system as star, so it’s two equations. And we had defined something called the determinant, ad minus bc. It was simply obtained from the coefficients of our linear system. And we had seen the proposition that when the determinant is non-zero, then the system star has a unique solution. And otherwise it has either no solutions or an infinity of solutions. We can now see geometrically why this is the case.
First of all, we can assume that a and b are not both 0. Otherwise, equation one is not really there. And similarly for c and d. So we look at equations one and equations two, and according to our previous discussion, we understand that these describe points that lie on two different lines in the plane. Now these lines in the plane are either parallel or not. It turns out, easy to see, that they are parallel precisely when the determinant is 0. So if the two lines are parallel, then either they are two distinct lines and they don’t meet, no solutions to the system star.
Or they are the same line, in which case all the points on that line satisfy the system star, infinitely many solutions. In the more interesting case where the lines are not parallel, then they must intersect at a single point and that is the unique solution of the system star, QED. We’re next going to look at systems of linear inequalities. And we’ll see that the situation is rather more different and really quite interesting.