### Video Transcript

The distance between π, five and one, one is five. What are the possible values of π?

So in this question, weβve been given information about the distance between a pair of coordinates. And so we recall the distance formula, which is a version of the Pythagorean theorem, tells us that the distance between two points π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two is π equals the square root of π₯ sub two minus π₯ sub one squared plus π¦ sub two minus π¦ sub one squared. And no, it doesnβt matter which coordinate we choose to be π₯ sub one, π¦ sub one and which we choose to be π₯ sub two, π¦ sub two. Weβll get the same result either way. Weβre just going to go in order here and let our first pair of coordinates be π, five and the coordinate π₯ sub two, π¦ sub two to be one, one.

Then, the distance between them is the square root of one minus π squared plus one minus five squared. But remember, we were told that the distance is equal to five. So instead, we can say that that expression on the right-hand side must be equal to five. So, how do we solve this equation for π? Well, letβs begin by squaring both sides. Five squared is 25. And on the right-hand side, we have one minus π squared plus one minus five squared. But of course, one minus five is equal to negative four, and then negative four squared is simply 16. So, we get 25 equals one minus π squared plus 16.

We now have two options at this stage. We could form and solve a quadratic equation by subtracting 25 from both sides, distributing the parentheses, and solving from there. Alternatively, letβs observe what happens if we subtract 16 from both sides. 25 minus 16 is nine, so we get nine equals one minus π squared. Next, we want to take the square root of both sides, but we remember when we do so, we have to take both the positive and negative square root of nine. So, the positive and negative square root of nine can be equal to one minus π. Well, the square root of nine is three, so we find that positive or negative three is equal to one minus π.

And now we can solve two separate equations for π. The first is the equation positive three equals one minus π, and the second is negative three equals one minus π. In both cases, letβs add π to both sides. We could alternatively at this stage multiply through by negative one. Essentially, weβre trying to make the sign of π positive just to make our life a bit easier. Our first equation becomes π plus three equals one and our second, π minus three equals one. Weβll subtract three from both sides of our first, giving us π equals negative two. And weβll add three to both sides of our second. And that gives us π equals four. And so, there are two possible values for π; π could be equal to negative two or π could be equal to four.

And it might seem unusual to have two possible solutions for π, but if we think about this geometrically, it makes a lot of sense. We could have a point with coordinates four, five and this lies at a distance of five units from point one, one. But then we also have this second point that is five units away from here and it has coordinates negative two, five. So, the possible values of π are negative two or four.