Video: Using Vectors to Find the Coordinates of a Vertex in a Square and Its Area

𝐴𝐡𝐢𝐷 is a square, in which the coordinates of the points 𝐴, 𝐡, and 𝐢 are (1, βˆ’8), (3, βˆ’10), and (5, βˆ’8). Use vectors to determine the coordinates of the point 𝐷 and the area of the square.

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Video Transcript

𝐴𝐡𝐢𝐷 is a square in which the coordinates of the points 𝐴, 𝐡, and 𝐢 are one, negative eight; three, negative 10; and five, negative eight. Use vectors to determine the coordinates of the point 𝐷 and the area of the square.

Firstly, it’s important to remember that we’re dealing with a square here. We know that, in a square, opposite sides are parallel and they’re 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. We’re also told that 𝐴, 𝐡, and 𝐢 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 𝐷, 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 𝐴 to 𝐷 and 𝐡 to 𝐢 must be parallel and equal in length.

So let’s begin by finding the vector 𝐡𝐢. One way to do this is to subtract the vector 𝑂𝐡 from the vector 𝑂𝐢. As a column vector, 𝑂𝐢 is five negative, eight and 𝑂𝐡 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 𝐡𝐢 is two, two.

This, in turn, means that the vector 𝐴𝐷 must also be two, two. And one way that we could travel from point 𝑂 to 𝐷 to help us find the vector 𝑂𝐷 would be to travel from 𝑂 to 𝐴 β€” that’s vector 𝑂𝐴 β€” and then travel from 𝐴 to 𝐷. So we’d add vector 𝐴𝐷. 𝑂𝐴 has vector one, negative eight. And we just found that vector 𝐴𝐷 is two, two. The sum of these is three, negative six. And we find that the vector 𝑂𝐷 is three, negative six. The coordinate of 𝐷 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, that’s the length of the line joining any two adjacent vertices. We’ll consider the vertices 𝐡 and 𝐢. We know that the vector 𝐡𝐢 is two, two. This means that the length of the line segment joining 𝐡 to 𝐢 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 that’s 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. It’s the square root of eight squared, which is of course eight. The area of the square is eight square units.

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