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coordinates

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6
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+1528

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 $n$?

Dec 10, 2023

#1
+222
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Let's analyze each piece of information given in the problem:

Distance from the x-axis: Since the point is 6 units below the x-axis, its y-coordinate is −6.

**Distance from the point (5,7): Using the distance formula, we have:(x−5)2+(y−7)2​=15

Distance from the origin: Using the distance formula again, we have:x2+y2​=n​

Since both x and y are negative, we can rewrite the equations with absolute values to avoid taking square roots of negative numbers:

Distance from the x-axis: ∣y∣=6⟹y=−6

**Distance from the point (5,7):(x−5)2+(y−7)2​=15⟹∣x−5∣2+∣y−7∣2=225

Distance from the origin:x2+y2​=n​⟹∣x∣2+∣y∣2=n

Now, substitute the value of y from the first point in the second and third equations:

**Distance from the point (5,7):∣x−5∣2+(−6−7)2=225⟹∣x−5∣2=126

Distance from the origin:∣x∣2+(−6)2=n⟹∣x∣2=n−36

Since ∣x−5∣2=126 and ∣x∣2=n−36, we have:

x2−10x+25=n−36

Solving for n, we get:

n=x2−10x+61

Since the equation for n involves only x, we can substitute the value of ∣x∣ from the equation for the distance from the origin:

n=(n−36)2−10(n−36)+61

Expanding and rearranging, we get a cubic equation:

n3−29n2−66n+1192=0

Solving this equation, we find that the only positive solution for n is:

n=784​

Dec 10, 2023