A point $(x,y)$ on the coordinate plane with both coordinates negative is a distance of 6 units from the $x$-axis. It is a distance of 15 units from the point $(8,3)$. It is a distance $\sqrt{n}$ from the origin. What is $n$?
To find the distance of a point from the $x$-axis, I need to consider the absolute value of its $y$-coordinate. So if the point is 6 units from the $x$-axis, that means $|y| = 6$. Since the point has both coordinates negative, I can write $y = -6$. To find the distance of a point from another point, I can use the distance formula, which is based on the Pythagorean theorem. So if the point is 15 units from $(8,3)$, that means $\sqrt{(x - 8)^2 + (y - 3)^2} = 15$. Substituting $y = -6$, I get $\sqrt{(x - 8)^2 + (-6 - 3)^2} = 15$, which simplifies to $\sqrt{(x - 8)^2 + 81} = 15$. Squaring both sides, I get $(x - 8)^2 + 81 = 225$, which implies $(x - 8)^2 = 144$. Taking the square root of both sides, I get $x - 8 = \pm 12$. So the point has coordinates $(-4,-6)$. To find the distance of the point from the origin, I can use the distance formula again. So the distance is $\sqrt{(-4)^2 + (-6)^2}$. Simplifying, I get $\sqrt{16 + 36}$, which is $\sqrt{52}$. Comparing this with the given distance $\sqrt{n}$, I see that $n = 52$. 