There are four points that are $5$ units from the line $y=13$ and $13$ units from the point $(7,13)$. What is the sum of the $x$- and $y$-coordinates of all four of these points?

Guest Apr 30, 2020

#1**+1 **

The sum of \(x\) and \(y\) coordinates of the four points is \(80\).

Since the distance between \((7,13)\) and a point is \(13\) units, and the distance from \(y = 13\) is \(5\), then it must be \(5\) units above or below the line. So, this must take the shape of a triangle, so we can use the pythagorean theroem to determine where the points lie. Since \(13^{2} = 169\), and \(5^{2} = 25\), \(169-25=144\). And the square root of \(144\) is \(12\), then we will have formed a triangle with base length \(12\) and height \(5\), to create a triangle with hypotenuse length \(13\). So, if we work it out, we are left with:

\(\left(-5,18\right) \left(-5,8\right) \left(19,18\right) \left(19,8\right)\)

And if we add them all up, we get \(80\).

- Keep working,

A fellow AoPSer.

Guest Jun 2, 2020