Question 677739: What is the point(s) that is a distance 7 from the point (2,5) and is on the line -x+y=3?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! -x+y=3
y=x+3
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d = sqrt((x2-x1)^2 + (y2-y1)^2)
d = sqrt((x-2)^2 + (y-5)^2)
d = sqrt((x-2)^2 + (x+3-5)^2)
d = sqrt((x-2)^2 + (x-2)^2)
d = sqrt(2(x-2)^2)
7 = sqrt(2(x-2)^2)
7^2 = 2(x-2)^2
49 = 2(x-2)^2
2(x-2)^2 = 49
(x-2)^2 = 49/2
x - 2 = sqrt(49/2) or x - 2 = -sqrt(49/2)
x - 2 = 7*sqrt(2)/2 or x - 2 = -7*sqrt(2)/2
x = 2 + 7*sqrt(2)/2 or x = 2 - 7*sqrt(2)/2
x = 6.949747 or x = -2.949747
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y = x+3
y = 6.949747+3
y = 9.949747
So when x = 6.949747, y is y = 9.949747. Therefore, the point (6.949747, 9.949747) is about 7 units away from (2,5) and this point lies on -x+y = 3
y = x+3
y = -2.949747+3
y = 0.050253
So when x = -2.949747, y is y = 0.050253. Therefore, the point (-2.949747, 0.050253) is also about 7 units away from (2,5) and this point lies on -x+y = 3
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