Find all points $(x,y)$ that are $5$ units away from the point $(2,7)$ and that lie on the line $y = 5x - 28.$

BRAlNBOLT May 24, 2024

#1**0 **

All points 5 units away from (2, 7) means that we can have a circle with radius 5 and center (2, 7).

Such a circle would look like \((x - 2)^2 + (y -7)^2 = 25 \).

Now we just have to find the intersection points of this circle and \(y = 5x - 28\).

This is pretty simple. We just have to find ordered pairs (x, y) that match for both equations. Luckily, we already isolated y from the second equation, so plugging in that y value into the first equation, we have

\((x -2)^2 + (5x -28 - 7)^2 = 25 \\ (x - 2)^2 + (5x - 35)^2 = 25\)

\(x^2 - 4x + 4 + 25x^2 - 350x + 1225 = 25 \\ 26x^2 - 354x + 1204 = 0\)

Solving from this, we get \(x=86/13, 7\).

Now plugging these values back into our second equation, we get \(y = 66/13, 7\).

So our two points are (86/13, 66/13) and (7, 7)!

Thanks! :)

NotThatSmart May 24, 2024