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The point $(x,y)$ in the coordinate has a distance of $6$ units from the $x$-axis, a distance of $15$ units from the point $(5,7)$, and a distance of $\sqrt{n}$ from the origin.  If both $x$ and $y$ are negative, what is $n$?

 Aug 26, 2023
 #1
avatar+36 
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The point (x,y) is in quadrant III, because x and y are both negative.

Because the point is 6 units from the x-axis, y is -6. So the point is (x,-6).

The distance from (x,-6) to (5,7) is 15. The difference in the y coordinates is 13, the distance is 15; find the difference in the x coordinates.
 

13+ b= 152

169 + b= 225

b2 = 56

b = √56

 

The point (x,y) is √56 units left of x=5, so the point is (5-sqrt(56)),-6)

The point is √n units from the origin:
(5 - √56)2 + 62 = (√n)2 = n

25 + 56 - 10√56 + 36 = n

n = 117 - 10√56 = 42.167 to 3 decimal places.

ANSWER: (approximately) 42.167

 Aug 27, 2023

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