Thread: Points Equidistant

In the xy plane, find all the points that are equidistant from (1, 2) and (3, 4) and
are also on the x axis.

(x-3)^2 + 4^2 = (x-1)^2 + 2^2

what would you with this next?

Perhaps you are in a course that is simply too advanced for you. You seem to be posting a number of questions in which you say you have "no idea" how to even begin. Do you know the "distance formula" for the distance between two points? If so you can write down the distance formula for the distance between (x, y) and (1, 2) or the distance between (x, y) and (3, 4). Since you are looking for points such that those two distances are the same, set them equal. That is exactly what Wilmer did for you- and, in fact, he squared both sides to get rid of the square root for you. Now you ask "what would you do with this next?" Didn't solve the equation occur to you?

One of the things you may have learned in geometry is that every point on the perpendicular bisector of a line segment is equal distance from the two endpoints of the line segment. So there will be a whole "line" of points equidistant from (1, 2) and (3, 4). You have the addtional requirement that the point (x, y) lies on the x-axis. What is true of all points on the x- axis? (Oops, I see that Wilmer has already included that in his equation!)

Pretty much everything except some simple algebra has been done for you. Solve that equation for x!

Would the answer then be 5? After I solved this equation this is what I have found. I isolated the one side and was left with 5 does this seem right?

Originally Posted by matgrl

Would the answer then be 5? After I solved this equation this is what I have found. I isolated the one side and was left with 5 does this seem right?

You can check this answer yourself. Substitute it back into the equation you were given. Does it work?

yes thank you