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3 years ago

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

Mathematics

1 answer:

kirza4 [7]3 years ago

7 0

<h3>Answers: x = 1 and x = 9</h3>

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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.

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

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.

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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. *​

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