### Video Transcript

Where must the coordinates of points πΆ be so that π΄π΅πΆπ· is a parallelogram? In that case, what is the area of the parallelogram?

In this question, weβre asked to make a parallelogram. We can remember that a parallelogram has two pairs of parallel and congruent sides. So if we look at our length π΄π΅, we can see that it must be five units long. Therefore, to draw a line from π· to πΆ, this must also be a horizontal line of length five units. We can confirm the coordinates of point πΆ by checking that the line π΅πΆ is parallel to the line π΄π·. We can see that on π΄π·, we moved three squares horizontally and seven squares vertically.

And to go from π΅ to πΆ, we can see that itβs also three squares horizontally and seven squares vertically. So line π΅πΆ is horizontal to π΄π· and confirms our point πΆ is that the coordinate six, five. Itβs worth noting at this point that our coordinate at negative four, five would also have created a parallelogram. However, we were told that our parallelogram was called π΄π΅πΆπ·. And in order to obey naming conventions, we list the vertices of a shape in the order of travel. If πΆ was to the left of π·, then the shape would have been called π΄π΅π·πΆ, which is not what we were told. And therefore, πΆ is not at negative four, five.

And now, for the second part of the question, using our πΆ coordinate, weβre asked for the area of the parallelogram. We can find the area of a parallelogram by multiplying the base times the vertical or perpendicular height. So using our formula then of base times height, we have our base of five units and a height of seven. So five times seven is 35.

So our final answer then is point πΆ is at coordinate six, five and the area is 35.