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The distance between π, five and one, one is five.
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What are the possible values of π?
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So in this question, weβve been given information about the distance between a pair of coordinates.
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And so we recall the distance formula, which is a version of the Pythagorean theorem, tells us that the distance between two points π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two is π equals the square root of π₯ sub two minus π₯ sub one squared plus π¦ sub two minus π¦ sub one squared.
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And no, it doesnβt matter which coordinate we choose to be π₯ sub one, π¦ sub one and which we choose to be π₯ sub two, π¦ sub two.
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Weβll get the same result either way.
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Weβre just going to go in order here and let our first pair of coordinates be π, five and the coordinate π₯ sub two, π¦ sub two to be one, one.
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Then, the distance between them is the square root of one minus π squared plus one minus five squared.
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But remember, we were told that the distance is equal to five.
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So instead, we can say that that expression on the right-hand side must be equal to five.
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So, how do we solve this equation for π?
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Well, letβs begin by squaring both sides.
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Five squared is 25.
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And on the right-hand side, we have one minus π squared plus one minus five squared.
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But of course, one minus five is equal to negative four, and then negative four squared is simply 16.
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So, we get 25 equals one minus π squared plus 16.
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We now have two options at this stage.
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We could form and solve a quadratic equation by subtracting 25 from both sides, distributing the parentheses, and solving from there.
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Alternatively, letβs observe what happens if we subtract 16 from both sides.
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25 minus 16 is nine, so we get nine equals one minus π squared.
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Next, we want to take the square root of both sides, but we remember when we do so, we have to take both the positive and negative square root of nine.
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So, the positive and negative square root of nine can be equal to one minus π.
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Well, the square root of nine is three, so we find that positive or negative three is equal to one minus π.
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And now we can solve two separate equations for π.
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The first is the equation positive three equals one minus π, and the second is negative three equals one minus π.
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In both cases, letβs add π to both sides.
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We could alternatively at this stage multiply through by negative one.
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Essentially, weβre trying to make the sign of π positive just to make our life a bit easier.
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Our first equation becomes π plus three equals one and our second, π minus three equals one.
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Weβll subtract three from both sides of our first, giving us π equals negative two.
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And weβll add three to both sides of our second.
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And that gives us π equals four.
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And so, there are two possible values for π; π could be equal to negative two or π could be equal to four.
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And it might seem unusual to have two possible solutions for π, but if we think about this geometrically, it makes a lot of sense.
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We could have a point with coordinates four, five and this lies at a distance of five units from point one, one.
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But then we also have this second point that is five units away from here and it has coordinates negative two, five.
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So, the possible values of π are negative two or four.