We know the `x` coordinate is 3, so our point is of the form `(3,y),` where we need to find `y` such that the distance between `(3,y)` and `(-2,-1)` is 13. According to the distance formula, the distance between these two points is given by
Since we want the distance to be 13, we solve
Now square both sides to get rid of the radical. We get
`169=25+(y+1)^2,` or `144=(y+1)^2.`
Here, we don't want to expand the right side. Instead, since `(y+1)^2=144,` we know that either
`y+1=12` or `y+1=-12.`
This gives the two solutions `y_1=11,` `y_2=-13,` so the two points whose distance from `(-2,-1)` is 13 are `(3,11)` and `(3,-13)`.
Here's the geometric view of the problem. The two points lie on the circle with center (-2,-1) and radius 13.