# Question Video: Finding the Unknowns in the Coordinates of a Point Using the Midpoint Formula Mathematics

Find the values of π and π so that (β2π, 2π + π) is the midpoint of the line segment between (β2, β3) and (2, 11).

02:50

### Video Transcript

Find the values of π and π so that the point negative two π, two π plus π is the midpoint of the line segment between negative two, negative three and two, 11.

In this question, we need to determine the values of two unknowns π and π. And these unknowns appear in the expressions for the coordinates of the midpoint of a line segment. And in particular, weβre given the endpoints of this line segment, the points negative two, negative three and two, 11. So to answer this question, letβs start by recalling how we determine the coordinates of a midpoint from its endpoints.

First, we recall the midpoint of a line segment is the point on the line segment equidistant from its two endpoints. And we can recall if the endpoints of the line segment have coordinates π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two, then the midpoint will have coordinates π₯ sub one plus π₯ sub two over two, π¦ sub one plus π¦ sub two over two. In other words, the π₯-coordinate of the midpoint of the line segment is the average of the π₯-coordinates of the endpoints. Similarly, the π¦-coordinate of the midpoint is the average of the π¦-coordinates of the endpoints.

So weβll start by setting π₯ sub one equal to negative two, π¦ sub one equal to negative three, π₯ sub two equal to two, and π¦ sub two equal to 11. Thatβs the coordinates of the endpoints given to us in the question. We then have π₯ sub one plus π₯ sub two over two needs to be equal to the π₯-coordinate of the midpoint. In our case, thatβs negative two π. Since π₯ sub one is negative two and π¦ sub one is two, we get that negative two π is equal to negative two plus two all over two. We can evaluate the right-hand side of this equation. The numerator is equal to zero. So the right-hand side of this equation is equal to zero. Therefore, negative two π is equal to zero.

We can then solve for π by dividing through by negative two. We get that π is equal to zero. We can now follow the same process for the π¦-coordinate. First, the π¦-coordinate of the midpoint is two π plus π. Next, in our formula, weβre told this is equal to the average of the π¦-coordinates of the endpoints. Thatβs π¦ sub one plus π¦ sub two all over two. Substituting in π¦ sub one is negative three and π¦ sub two is equal to 11, we get that this is equal to negative three plus 11 all over two.

And now we can simplify. First, our value of π is equal to zero. So we can substitute π is equal to zero into this equation. And since two times zero is zero, the left-hand side simplifies to give us π. We can then evaluate the right-hand side of this equation. First, in the numerator, we have negative three plus 11, which is equal to eight, which then gives us π is equal to eight divided by two, which we can calculate is equal to four. Therefore, we were able to show if the point negative two π, two π plus π is the midpoint of the line segment between negative two, negative three, two, 11, then π must be equal to zero and π must be equal to four.