### 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, weβre given the endpoints of this line segment, the points negative two, negative three and two, 11. So to answer this question, letβs 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 weβll 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. Thatβs 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, thatβs 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, weβre told this is equal to the average of the π¦-coordinates of the endpoints. Thatβs 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.