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5. The distance between two given points (5,2) and (x,5) is 5 units. Find the possible values of x. *​

User Thammas
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1 Answer

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Answers: x = 1 and x = 9

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Step-by-step explanation:

We'll use the distance formula here. Rather than compute the distance d based on two points given, we'll go in reverse to use the given distance d to find what the coordinate must be to satisfy the conditions.

We're given that d = 5

The first point is
(x_1,y_1) = (5,2) and the second point has coordinates of
(x_2,y_2) = (x,5) where x is some real number.

We'll plug all this into the distance formula and solve for x.


d = √((x_1-x_2)^2+(y_1-y_2)^2)\\\\5 = √((5-x)^2+(2-5)^2)\\\\5 = √((5-x)^2+(-3)^2)\\\\5 = √((5-x)^2+9)\\\\√((5-x)^2+9) = 5\\\\(5-x)^2+9 = 5^2\\\\(5-x)^2+9 = 25\\\\(5-x)^2 = 25-9\\\\(5-x)^2 = 16\\\\5-x = \pm√(16)\\\\5-x = 4 \ \text{ or } \ 5-x = -4\\\\-x = 4-5 \ \text{ or } \ -x = -4-5\\\\-x = -1 \ \text{ or } \ -x = -9\\\\x = 1 \ \text{ or } \ x = 9\\\\

This means that if we had these three points

  • A = (5, 2)
  • B = (1, 5)
  • C = (9, 5)

Then segments AB and AC are each 5 units long.

5. The distance between two given points (5,2) and (x,5) is 5 units. Find the possible-example-1
User Waclock
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