Finally! A use for the point-slope form of linear equations!

I’m a slope-intercept man, myself.

y=mx+b, baby.

That’s what you really need in life, right? The slope is the rate, the y-intercept is the starting value. What more information could be necessary?

But the other day I needed the point-slope form. Here’s why.

The textbook I am using for College Algebra introduces difference quotients at the end of a section on operations on functions. It doesn’t flow and I don’t get why it’s there, but it is. So I want to give my students the best possible chance of making sense of the topic.

So I decided to cook up a dynamic graphical demonstration using the software Fathom. I have written about other uses of Fathom elsewhere, notably in an article about Fathom’s power for investigating experimental probabilities. In this case, I was using it to graph a parabola and a secant line to the parabola so that I could move one of the two intersection points back and forth while leaving the other point fixed. I arbitrarily chose (1,1) for the fixed point. The video below is a crummy little screen capture, but it gives the flavor of what I wanted to do.

The function itself (the parabola) was easy enough to plot, and I could use the slider to move the intersection point back and forth on the curve. But how was I going to define that line?

The  slider works by giving a value to a variable called V1. So I wanted the line that goes through (1,1) and (V1, V1^2). (NOTE: by V1^2, I mean “the square of V1”- I’ll figure out how to put exponents in text another time)

Slope intercept form assumes that I know the slope (which is going to change as the point moves) and that I know y-intercept (which is also going to change as the point moves). Indeed figuring out the y-intercept seemed like it was going to be a hassle.

Enter the point-slope form, which is:

Given the two points named above, the slope is:

Interestingly, this slope simplifies:

So the equation for the line is:

But I need to solve for y in order to tell Fathom how to plot the line, so a bit of solving yields this:

And this is good enough for what I needed to do. It’s the equation that underlies the green line in the demo video above. But by working a little bit more, I can get it into slope-intercept form:

I don’t have time to try, but I sense that this simple equation would have been harder to derive by starting with the y-intercept form.

Advertisement