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Distance Between Two Points

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Sometimes, the test gives two points and asks us for the distance between them. Other times, we will need to find distance in the x-y plane in the course of answering a geometric question. Something like is the triangle isosceles, something like that. Or is the figure a square? Big Idea number 1.

In the x-y plane, horizontal and vertical distances are very easy to find. If two points are on the same horizontal line, all we have to do is subtract the, the x-coordinates. We subtract bigger minus smaller because distance is always positive. If two points are on the same vertical line, we subtract the y-coordinates. Incredibly easy.

We can't do anything this simple if the two points are separated by a, a slanted line. Let's think about two points diagonally separated. So for example, suppose we have this. So we can't just do any kind of subtraction, but think about it this way.

If we draw the slope triangle we were using to find the slope, remember the slope triangle, suddenly things become much clearer. Okay so here are the same points again but now we've drawn a slope triangle. Well it's very easy to find the rise and the run on that slope triangle. The rise is three. We can see there's a, the vertical leg has a distance of 3, a length of 3.

The horizontal leg, the run, has a, a length of 4. So if the horizontal leg is 4 and the vertical leg is 3, this is just a 3-4-5 triangle and so the distance between those two points the hypotenuse has to be 5. Now it won't always be the case that the three sides of a triangle are nice neat Pythagorean triplets, although this is quite common on the test. Nevertheless, it will always be the case that the slope triangle is a right triangle, by definition the slope triangle is a right triangle.

The horizontal and vertical legs have lengths that are easy to find, and the length of the diagonal will always be the hypotenuse of the sloped triangle. We can always find that length by the Pythagorean theorem so we're using the Pythagorean theorem to find the distance between any two points. That's Big Idea number 2. Here's a practice problem.

Find the distance between those two points, pause the video and then we'll talk about this. Okay. So if we find the x-leg, the run between them, we get a run of 8 and a rise of 4, a y-leg of 4.

We could do the Pythagorean Theorem with legs of 4 and 8 but it's easier if we scale down first by the greatest common factor, a scale factor of 4. We scale down to a much smaller triangle, with legs of 1 and 2. Well that's very easy to solve. If we have legs of 1 and 2, we square those, we get c squared equals 5, c equals the square root of 5.

Now we'll scale back up by a scale factor of 4. And we get a hypotenuse on the larger sloped triangle of 4 root 5, and that's the distance between those two points. So notice that we could use proportional thinking to really simplify that calculation. Now there is a formula known as the distance formula.

I absolutely refuse to teach it, neither of you know it, I encourage you not to use it. I think sticking to this formula is actually ultimately a more time consuming way that impedes understanding and you understand much more deeply if you think about it the way I'm showing in this video using the Pythagorean theorem. So if you use the Pythagorean theorem to find the hypotenuse of a sloped triangle, you're thinking visually and you, you can also use the proportional thinking to simplify your calculation.

Of course, using the slope formula would, the distance formula would not allow you to do that. So there are a variety of reasons that using the formula is an exceptionally bad idea and it's much better to think about this situation carefully. When you do think visually about the x-y plane, you understand much more deeply than when you simply plug in to a memorized formula.

Now we can talk about circles in the x-y plane. Of course, according to its fundamental definition, a circle is the set of all points equidistant from a fixed center. Everything we said about distance comes into play with circles in the x-y plane. Here's a practice problem, pause the video and then we'll talk about this.

Okay. A circle in the y, x-y plane has a center of 6, 3 and a radius of 5. Find the two x-intercepts. Think about the two radii that go from the center to these two x-intercepts.

Each radius has a length of 5, and it's the hypotenuse of a slope triangle. Also, the point 6, 3 is clearly 3 units above the x-axis. So let's think about all this visually. So we have those two little triangles there, each one has a hypotenuse of 5, and it's a distance of 3 above the x-axis so that vertical length is 3.

Well of course what we have there are two 3-4-5 triangles, so the horizontal lengths each have a length of 4. And so we, starting at 6, we go 4 to the right and 4 to the left, we get two inter, x-intercepts of 2, 0 and 10, 0 and those are the x-intercepts. If the circle is centered at the origin, then the slope triangle of each radius has a horizontal leg of the absolute value of x and a vertical leg of the absolute value of y, regardless of what x and y are.

Thus, the equation for a circle with center 0, 0 and radius r must be, x squared plus y squared equals r squared. So notice that we're really just doing the Pythagorean theorem, define the equation of a circle. So horizontal and vertical distances in the x-y plane, we just use subtraction. Very easy.

For slanted distances, we draw or imagine the slope triangle and use the Pythagorean theorem, the distance between the two points is the hypotenuse ff the slope triangle. Remember Pythagorean triplets, remember to use scale factors to simplify your calculations, all those things we talked about back in the geometry lessons. For a circle in the x-y plane, each slanted radius forms a slope triangle and all these slope triangles have equal hypotenuses because the hypotenuse is the radius.

The equation of a circle with radius r centered at the origin is x squared plus y squared equals r squared.

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