+415 Find all points $(x,y)$ that are $5$ units away from the point $(2,7)$ and that lie on the line $y = 5x - 28.$
+1365 If we drew the graph for all the points 5 units away from (2, 7), we would get a circle with radius \(\sqrt5\) and center (2, 7).
A circle like that would have the equation \((x - 2)^2 + (y -7)^2 = 25\).
Now, we just have to find the intersection points with the graph of \(y = 5x - 28\).
We get a system of equations like this, and we just have to find (x, y). Luckily, y is already isolated in the second equation, so we just have to sub that into the first equation. We get
\((x -2)^2 + (5x -28 - 7)^2 = 25 \\ (x - 2)^2 + (5x - 35)^2 = 25\)
\(x^2 - 4x + 4 + 25x^2 - 350x + 1225 = 25 \\ 26x^2 - 354x + 1204 = 0\)
From this, we get\(x=86/13, 7\)
Plugging these back for y, we get \(y = 66/13, 7\).
This means our two points are \((86/13, 66/13) & (7, 7)\)
Thanks! :)
