### Video Transcript

π΄π΅πΆπ· is a square in which the coordinates of the points π΄, π΅, and πΆ are one, negative eight; three, negative 10; and five, negative eight. Use vectors to determine the coordinates of the point π· and the area of the square.

Firstly, itβs important to remember that weβre dealing with a square here. We know that, in a square, opposite sides are parallel and theyβre equal in length. We also know that adjacent sides are at a right angle to one another, although that may or may not be useful as we move through this question. Weβre also told that π΄, π΅, and πΆ are the points one, negative eight; three, negative 10; and five, negative eight, respectively. And so we might choose to plot these points on a coordinate grid as shown.

It follows that π·, the fourth vertex of the square, would be somewhere up here. And in fact, there are a couple of ways that we could use vectors to work out the exact coordinates of this point. Now it may not look like it on my diagram, but we do know that the vectors joining π΄ to π· and π΅ to πΆ must be parallel and equal in length.

So letβs begin by finding the vector π΅πΆ. One way to do this is to subtract the vector ππ΅ from the vector ππΆ. As a column vector, ππΆ is five negative, eight and ππ΅ is three, negative 10. To subtract vectors, we simply subtract their individual components. Five minus three is two. And minus eight minus negative 10 is also two. So we see that the vector π΅πΆ is two, two.

This, in turn, means that the vector π΄π· must also be two, two. And one way that we could travel from point π to π· to help us find the vector ππ· would be to travel from π to π΄ β thatβs vector ππ΄ β and then travel from π΄ to π·. So weβd add vector π΄π·. ππ΄ has vector one, negative eight. And we just found that vector π΄π· is two, two. The sum of these is three, negative six. And we find that the vector ππ· is three, negative six. The coordinate of π· therefore must be three, negative six.

The second part of this question asks us to find the area of the square. So we recall that, to find the area of the square, we simply square its width or its height. So how do we find the width or the height of our square? Well, thatβs the length of the line joining any two adjacent vertices. Weβll consider the vertices π΅ and πΆ. We know that the vector π΅πΆ is two, two. This means that the length of the line segment joining π΅ to πΆ is the magnitude of this vector. And we find the magnitude of the vector by finding the square root of the sum of the squares of each of its components.

So thatβs the square root of two squared plus two squared, which is the square root of eight. And we see that each side of our square must therefore be the square root of eight units. The area is the square of this value. Itβs the square root of eight squared, which is of course eight. The area of the square is eight square units.