Find the distance and midpoint for the following two points.  (5,8), (-3,2)

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txmedteach | High School Teacher | (Level 3) Associate Educator

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To start, we need to think of what distance and midpoint mean in this context. Let's graph these two points to see what we're looking at.

To find the midpoint is not that bad, actually! We simply take the average x value and average y value! Here's the formula:

`M = ((x_1+x_2)/2, (y_1+y_2)/2)`

All we need to do here is substitute our x and y values into the formula:

`M = ((5+(-3))/2, (8+2)/2)`

Now, we simplify to get:

`M = (1, 5)`

Let's graph the point to see if we're right.

Well, that certainly looks like the midpoint to me! If you want to check, you can see that the midpoint is 3 squares above and 4 squares to the right of (-3, 2), and to get to (5,8) from our midpoint, we need to move the same 3 up and 4 to the right. It is confirmed!

Now, to find the distance. The distance formula proposed is useful (shown here):

`D = sqrt((y_2-y_1)^2 + (x_2-x_1)^2)`

However, this doesn't really hit at the basis for how we find distance. If you wanted to find distance, you can treat your points like points on a right triangle. To complete the triangle, make a vertical line from one point, and a horizontal line from another:

Recognize that the hypotenuse of this right triangle is also the distance between the two points! Also, recall the Pythagorean Theorem:

`a^2 + b^2 = c^2`

So, considering that our hypotenuse is c, all we need to do is find the length of the legs to solve this equation! If you look at the graph, you'll easily see that the leg lengths are 8 for the horizontal leg and 6 for the vertical leg.

In the distance formula, this is represented by the differences between x and y terms, so you'll see this way is no different!

`x_2-x_1 = -3-5 = -8`

`y_2 - y_1 = 2-8 = -6`

Well, -6 is different from 6, same with -8 and 8. However, since we're taking the square of the values, there isn't a difference!

We'll put these values into the Pythagorean Theorem/Distance Formula:

`64 + 36 = c^2`

`100 = c^2`

Now, we take the square root:

`10 = c`

So our distance will be 10. Another way to see this is that the triangle we created on the graph is a 3,4,5 right triangle (just doubled).

I hope that helps!


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