Question Video: Finding the Coordinates of Points Using the Midpoint Formula | Nagwa Question Video: Finding the Coordinates of Points Using the Midpoint Formula | Nagwa

Question Video: Finding the Coordinates of Points Using the Midpoint Formula Mathematics • Third Year of Preparatory School

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Suppose 𝐴(βˆ’7, βˆ’4), 𝐡(6, βˆ’9), and 𝐷(8, βˆ’2). If 𝐢 is the midpoint of both line segment 𝐴𝐡 and line segment 𝐷𝐸, find 𝐸.

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Video Transcript

Suppose 𝐴 is negative seven, negative four; 𝐡 is six, negative nine; and 𝐷 is eight, negative two. If 𝐢 is the midpoint of both line segment 𝐴𝐡 and line segment 𝐷𝐸, find point 𝐸.

Let’s first sketch what we know. We have a line segment 𝐴𝐡 with the midpoint 𝐢. And 𝐢 is also the midpoint of line segment 𝐷𝐸. We’re given the coordinates for 𝐴, 𝐡, and 𝐷. Our end goal is to find the coordinates of point 𝐸. But before we can find 𝐸, we’ll need to know 𝐢. Once we find 𝐢, we can find 𝐸. And to do both of these things, we’ll need to remember that the midpoint formula looks like this.

The π‘₯-coordinate of the midpoint is found by taking the π‘₯-coordinates from the endpoints and dividing by two. And the 𝑦-coordinate of the midpoint is found by averaging the 𝑦-coordinates of the two endpoints. Since 𝐢 is the midpoint of 𝐴 and 𝐡, we’ll let 𝐴 be π‘₯ one, 𝑦 one and 𝐡 be π‘₯ two, 𝑦 two. The midpoint 𝐢 will be located at negative seven plus six over two, negative four plus negative nine over two. Negative seven plus six over two is negative one-half. And negative four plus negative nine is negative 13. So, the 𝑦-coordinate is negative 13 over two. Now, we know where 𝐢 is located. And we’re ready to think about 𝐸.

If 𝐢 is also the midpoint of 𝐷𝐸, then the coordinates of 𝐢 will be equal to the π‘₯-coordinates of 𝐷 and 𝐸 averaged together and the 𝑦-coordinates of 𝐷 and 𝐸 averaged together. We’re given the coordinates of 𝐷. That’s eight, negative two. And so, we plug that in. From here, we’ll make two separate equations. We’ll set negative one-half equal to eight plus the π‘₯-coordinate of 𝐸 over two. And negative 13 over two is equal to negative two plus the 𝑦-coordinate of 𝐸 over two. We’ll give ourselves a little bit more room.

Since all of the denominators are two, then the numerators are equal to each other. Negative one equals eight plus the π‘₯-coordinate of 𝐸. And to solve for that missing value, we subtract eight from both sides. And we see the π‘₯-coordinate for point 𝐸 is negative nine. To solve for the 𝑦-coordinate of point 𝐸, we add two to both sides. The 𝑦-coordinate of 𝐸 is negative 11. In coordinate form, point 𝐸 is located at negative nine, negative 11.

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