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Question 1206193: Given (20,8) and (x,−1), find all x such that the distance between these two points is 15. Separate multiple answers with a comma.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! formula for distance between two point is sqrt((x1-x2)^2 + (y1-y2)^2).
(x1,y1) = (20,8)
x2,y2) = (x,-1)
distance between the two points is 15.
formula becomes sqrt((20-x)^2 + (8--1)^2) = 15.
square both sides of this equation to get (20-x)^2 + (8--1)^2 = 225.
simplify to get 400 -40x + x^2 + 81 = 225.
combine like terms and rearrange the equation in descending order of degree to get x^2 - 40x + 481 = 225.
subtract 225 from both sides of the equation to get x^2 - 40x + 256 = 0.
factor this quadratic equation to get (x-8) * (x-32) = 0.
solve for x to get x = 32 or x = 8.
when x = 8, sqrt((20-x)^2 + (8--1)^2) = 15 becomes sqrt((20-8)^2 + 81) = 15 which becomes sqrt(12^2 + 81) = 15 which becomes sqrt(144 + 81 = 15 which becomes sqrt(225) = 15 which is true.
when x = 32, sqrt((20-x)^2 + (8--1)^2) = 15 becomes sqrt((20-32)^2 + 81) = 15 which becomes sqrt((-12)^2 + 81) = 15 which becomes sqrt(144 + 81) = 15 which becomes sqrt(225) = 15 which is also true.
your solution is that x = 32 or x = 8.
you can also show as x = 32,8.
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