How would I find a point on the y-axis equidistant from two other points?
The points are$$(5,-5) and (1,1)$$
I tried doing this visually and came up with (0,-5). This wasn't correct once I applied the distance formula to check the distance between that point and the two others.
$\endgroup$ 32 Answers
$\begingroup$You want the point where the perpendicular bisector of the two points cuts the $y$-axis.
The slope of the line between $(5,-5)$ and $(1,1)$ is $-\frac{3}{2}$, so the slope of the normal is $\frac23$. Hence the normal through the mid-point $(3,-2)$ cuts the $y$-axis at $(0,-4)$.
$\endgroup$ 4 $\begingroup$A point on the $y-$axis is of the form $(0,y)$.
The distance between $(0,y)$ and $(5,-5)$ is: $$\sqrt{(5-0)^2+(-5-y)^2}=\sqrt{25+(5+y)^2}$$
The distance between $(0,y)$ and $(1,1)$ is: $$\sqrt{(1-0)^2+(1-y)^2}=\sqrt{1+(1-y)^2}$$
The two distances are equal, so also their squares:
$$25+(5+y)^2=1+(1-y)^2$$
Now you have to solve for $y$.
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