The points on $y$ -axis

As points (0,q) on the y-axis lie on the perpendicular bisectors of the given lines, we equate the squares of the distances from them. $\frac{(q-10)^2}{5} = \frac {(q-4)^2}{5/4}$ $\rightarrow q^2 - 20q + 100 = 4q^2 - 32q + 64$ $\rightarrow 3q^2 - 12q -36=0$ $\rightarrow 3(q-6)(q+2)=0$ (0,-2) and (0,6) are the points that are equidistant from the given lines and hence on the perpendicular bisectors of the lines. The distance between these two points is equal to 8.

Let the angle of the line $y=\frac{1}{2}x+4$ be $a$ and the angle of $y=2x+10$ be $b$ .

The slope of the first line is the reciprocal of the second line's slope, which means that $a=\pi - b$ .

Each point, that same the same distance to both lines, must lie on either bisector of the lines (there are two of them). The angle of one bisector is the arithmetic average of the line's angles, which is $\frac{a+b}{2}=\frac{a+\pi-a}{2}=\pi/2$ . This means that the slope of this bisector is $k_1=1$ . The other bisector's angle is perpendicular to the first one's, so it's slope must be $k$ , where $1\times k=-1$ , which means $k_2=-1$ .

Now let's find the point where the lines intersect:

$y=2x+10$

$y=\frac{1}{2}x+4$

$2x+10=\frac{1}{2}x+4$

$\frac{3}{2}x=-6$

$x=-4$

$y=2x+10=2\times -4 +10=2$

So the lines intersect at $(-4,2)$ , which means both bisectors must go through this point.

Now, let's find the equations of the bisectors:

$y-y_0=k_1(x-x_0)$

$y-2=1(x+4)$

$y=x+6$

$y-y_0=k_2(x-x_0)$

$y-2=-1(x+4)$

$y=-x-2$

By replacing the x with 0 in both equations, it appears that the lines intersect the y-axis at $(0,6)$ and $(0,-2)$ . Their distance is $\sqrt{(6-(-2))^2+(0-0)^2}=\boxed{8}$ .