Find the point on the y-axis which is equidistant from the points $(6, 2)$ and $ (2, 10)$. Find the point on the y-axis which is equidistant from the points $(6, 2) $ and $ (2, 10)$.
Please help, there are no examples of this kind of sum in my book! I don't know how to solve it.
 A: Find the locus of all the points that are equidistant from the points $(6, 2)$ and $(2, 10)$, that is, the line that passes through the middle point of $(6, 2)$ and $(2, 10)$. All the points of that line are equidistant from both points.
You are looking for the point that satisfies two condition: (1) It's equidistant from both points and (2) it's in the Y axis (the x=0 line). Therefore, the point you are looking for is the intersection of the two lines.
A: HINT:
Any point on the $y$ axis can be stated as $P(0,b)$
Do you know how to calculate the distance between two points in $2$D?  
A: The locus of all points plane equidistant to two distinct points $P = (x_1,y_1)$, $Q = (x_2, y_2)$ is the line $\ell$ that bisects the segment $\overline{PQ}$ perpendicularly.  This line $\ell$ is given by
$$\ell: \;\; y = y_0 + (x - x_0) \cdot \left(-\frac{x_2-x_1}{y_2-y_1}\right) \;,$$
where $(x_0, y_0)$ is the midpoint of $\overline{PQ}$:
$$(x_0, y_0) = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \;.$$
The point you want is the intersection of $\ell$ and the $y$-axis.  The $x$-coordinate of this point is $0$.  You get the $y$-coordinate by plugging this value of $x$ into the equation for $\ell$ given earlier:
$$\left(0, y_0 + x_0 \cdot \frac{x_2-x_1}{y_2-y_1} \right)\;,$$
...or, after plugging in the expression for $(x_0, y_0)$,
$$\left(0, \frac{x_2^2-x_1^2 + y_2^2 - y_1^2}{2 (y_2-y_1)} \right)\;.$$
