WEBVTT
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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.
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Use vectors to determine the coordinates of the point π· and the area of the square.
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Firstly, itβs important to remember that weβre dealing with a square here.
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We know that, in a square, opposite sides are parallel and theyβre equal in length.
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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.
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Weβre also told that π΄, π΅, and πΆ are the points one, negative eight; three, negative 10; and five, negative eight, respectively.
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And so we might choose to plot these points on a coordinate grid as shown.
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It follows that π·, the fourth vertex of the square, would be somewhere up here.
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And in fact, there are a couple of ways that we could use vectors to work out the exact coordinates of this point.
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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.
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So letβs begin by finding the vector π΅πΆ.
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One way to do this is to subtract the vector ππ΅ from the vector ππΆ.
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As a column vector, ππΆ is five negative, eight and ππ΅ is three, negative 10.
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To subtract vectors, we simply subtract their individual components.
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Five minus three is two.
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And minus eight minus negative 10 is also two.
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So we see that the vector π΅πΆ is two, two.
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This, in turn, means that the vector π΄π· must also be two, two.
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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·.
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So weβd add vector π΄π·.
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ππ΄ has vector one, negative eight.
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And we just found that vector π΄π· is two, two.
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The sum of these is three, negative six.
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And we find that the vector ππ· is three, negative six.
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The coordinate of π· therefore must be three, negative six.
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The second part of this question asks us to find the area of the square.
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So we recall that, to find the area of the square, we simply square its width or its height.
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So how do we find the width or the height of our square?
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Well, thatβs the length of the line joining any two adjacent vertices.
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Weβll consider the vertices π΅ and πΆ.
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We know that the vector π΅πΆ is two, two.
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This means that the length of the line segment joining π΅ to πΆ is the magnitude of this vector.
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And we find the magnitude of the vector by finding the square root of the sum of the squares of each of its components.
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So thatβs the square root of two squared plus two squared, which is the square root of eight.
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And we see that each side of our square must therefore be the square root of eight units.
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The area is the square of this value.
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Itβs the square root of eight squared, which is of course eight.
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The area of the square is eight square units.