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distance on the plane

0
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A point (x, y) is a distance of 6 units from the x-axis. It is a distance of 5 units from the point (8, 3). It is a distance \sqrt(n) from the origin. Given that x<8, what is n?

Sep 25, 2019

#1
+109740
+1

Using the Pythgorean Theorem  the point  can be  represented  as :

[x, y ] =  [sqrt (n - 6^2), 6 ]

So we have that the distance from (x,y) to (8,3)  =  5 units....so......

[ sqrt (n - 6^2) - 8]^2   +  [ 6 -3]^2  = 25      simplify

(n - 36)  - 16sqrt(n - 36) + 64  + 9  = 25

n - 36 + 73 - 25  =  16sqrt(n - 36)

n + 12  =  16sqrt(n - 36)      square both sides

n^2 + 24n + 144 = 256n - 9216

n^2 - 232n + 9360  = 0   factor as

(n - 52) (n - 180)  = 0

If n = 52   then x  =  sqrt (52 - 36)  =  sqrt(16)  =  4

If n = 180  then x  = sqrt (180 - 36)  = sqrt (144)  = 12

So  (x , y )  =  ( 4, 6)

And n = 52

Here's a graph

Sep 25, 2019
edited by CPhill  Sep 25, 2019
edited by CPhill  Sep 25, 2019
edited by CPhill  Sep 26, 2019
edited by CPhill  Sep 26, 2019
edited by CPhill  Sep 26, 2019
edited by CPhill  Sep 26, 2019