# Find all the points having an x-coordinate of 3 whose distance from the point (-2, -1) is 13

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### 1 Answer

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

`sqrt((3-(-2))^2+(y-(-1))^2)=sqrt(5^2+(y+1)^2)=sqrt(25+(y+1)^2).`

Since we want the distance to be 13, we solve

`13=sqrt(25+(y+1)^2).`

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.

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