### Video Transcript

π΄π΅πΆπ· is a parallelogram where the coordinates of the points π΄, π΅, and πΆ are four, negative nine; three, two; and five, five, respectively. Find coordinates of vertex π·.

Since we know that π΄π΅πΆπ· is a parallelogram, we know that opposite sides are parallel. While it is possible to solve this without graphing, it still might be a good idea to sketch a graph, especially if youβre new to these type of problems. Once you sketch the coordinate system, we add the point four, negative nine which is π΄; three, two which is π΅; and five, five which is πΆ. We can connect these three points.

Now, since we know that the opposite sides are parallel and we know that parallel lines have the same slope, we can find the slope of line segment π΅πΆ. And using that slope, we can find the missing point π·. Our slope is the changes in π¦ over the changes in π₯. We sometimes call it the rise over the run. We want to know from π΅ to πΆ, what are the changes in π¦? From two to five is a rise of three. And then if we look at our changes in π₯, from three to five is a change of positive two. That means the slope between these two points is three over two.

It also means from point π΄, we need to go up three and right two to find point π·. Since we started at negative nine with point π΄, we need to go up three. And from positive four, weβll go right two for our point π·. The new coordinate should be found at positive six, negative six. And then we could connect the points πΆ and π· for our parallelogram.

If we wanted to check and make sure this was true, we could check the slope of the other two lines. To go from π΄ to π΅, we go from negative nine up to two. Thatβs a rise of positive 11. But from π΄ to π΅, weβre moving left one, so we have a run of negative one. This means the slope from π΄ to π΅ is negative 11. This means if we start at πΆ, the point five, five, and we go down 11, right one, we should arrive at π·.

Down 11 and right one, we arrive at the point six, negative six. So we can say that π· is located at six, negative six.