### Video Transcript

Given that the point πΆ negative one-half, zero, negative two is the midpoint of the line segment ππ, where the coordinates of π΄ and π΅ are π plus five, eight, π plus four and negative six, π plus seven, five, respectively, what is π plus π minus π?

In this question, weβre given the coordinates of a point πΆ, and weβre told that πΆ is the midpoint of the line segment ππ, and weβre given the coordinates of point π΄ and point π΅. However, these coordinates include the unknowns π, π, and π. We need to use the fact that πΆ is the midpoint of the line segment ππ to determine the value of π plus π minus π. So to do this, letβs start by recalling how we find the midpoint of a line segment. We recall that we can find the midpoint of the two points π₯ one, π¦ one, π§ one and π₯ two, π¦ two, π§ two by taking the average of their coordinates. The midpoint of these two points is given by π₯ one plus π₯ two all over two, π¦ one plus π¦ two all over two, π§ one plus π§ two all over two.

And itβs worth reiterating here if we have a line segment defined by these two points as the endpoints, then this would also give us the midpoint of this line segment, so we can use this to answer our question. π₯ one, π¦ one, π§ one will be our point π΄, π₯ two, π¦ two, π§ two will be our point π΅, and then the midpoint of these will be our point πΆ. So letβs apply this formula on the two points given to us in the question. First, letβs find the π₯-coordinate of our midpoint. We need to take the average of the π₯-coordinates of our endpoints. That gives us π plus five plus negative six all divided by two. And of course, this gives us the π₯-coordinate of the midpoint of this line segment. But we know this is equal to the π₯-coordinate of point πΆ, which is negative one-half.

So this gives us an equation: π plus five plus negative six all over two is equal to negative one-half. And we can solve this equation for π. Weβll start by multiplying through by two. On the left-hand side, we get π plus five plus negative six, which we can simplify to give us π minus one. And on the right-hand side of this equation, we have negative one-half multiplied by two, which is equal to negative one. Then we can solve for π by just adding one to both sides of our equation. We get that π has to be equal to zero.

This isnβt the only thing we can do, though; we can do exactly the same for the π¦-coordinate. Taking the average of the π¦-coordinates of our end points, we get eight plus π plus seven all over two. And of course, this has to be equal to the π¦-coordinate of our midpoint, which is the π¦-coordinate of point πΆ, which we know is zero. So we now get another equation for π. We have eight plus π plus seven all over two is equal to zero.

Now we could rearrange and solve for our value of π. However, for this equation to be true, the numerator must be equal to zero. In other words, π is going to have to be equal to negative 15. Weβll do this one more time to find our value of π. We take the average of the two π§-coordinates of our endpoints, and this is going to have to be equal to the π§-coordinate of our midpoint πΆ, which is negative two. So this gives us the equation π plus four plus five all over two is equal to negative two. And now we can solve this equation to find the value of π. Weβll start by multiplying through by two.

On the left-hand side, we have π plus four plus five, which simplifies to give us π plus nine. And on the right-hand side of our equation, we have negative two times two, which is negative four. And finally, we can solve for our value of π by subtracting nine from both sides of the equation. We get that π is equal to negative 13. And now that we found the values of π, π, and π, we can just substitute them into the expression weβre asked to find. Substituting π is equal to zero, π is equal to negative 13, and π is equal to negative 15, we have π plus π minus π is equal to zero plus negative 13 minus negative 15. And we can just calculate this expression. Subtracting negative 15 is the same as adding 15, so we get negative 13 plus 15 which is equal to two.

Therefore, we were able to show if the point negative one-half, zero, negative two is the midpoint of the line segment with endpoints π plus five, eight, π plus four and negative six, π plus seven, five, then we must have that π plus π minus π is equal to two.