Perpendicular Bisector of Made from Two Points For a National Board Exam Review: 

Find the equation of the perpendicular bisector of the line joining
  (4,0) and (-6, -3)

Answer is 20x + 6y + 29 = 0
I dont know where I went wrong. This is supposed to be very easy:
Find slope between two points:
$${ m=\frac{y^2 - y^1}{ x^2 - x^1 } = \frac{-3-0}{-6-4} = \frac{3}{10}}$$
Obtain Negative Reciprocal:
$${ m'=\frac{-10}{3}}$$
Get Midpoint fox X
$${ \frac{-6-4}{2} = -5 }$$
Get Midpoint for Y
$${ \frac{-0--3}{2} = \frac{3}{2} }$$
Make Point Slope Form: 
$${ y = m'x +b = \frac{-10}{3}x + b}$$
Plugin Midpoints in Point Slope Form
$${ \frac{3}{2} = \frac{-10}{3}(-5) + b}$$
Evaluate b
$${ b = \frac{109}{6}}$$
Get Equation and Simplify
$${ y = \frac{-10}{3}x + \frac{109}{6}}$$
$${ 6y + 20x - 109 = 0 }$$
Is the problem set wrong? What am I doing wrong?

Just for variety, a different approach:
If a point $(x,y)$ is on the perpendicular bisector, then it is equidistant from $(4,0)$ and $(-6,-3)$, so
$$
(x-4)^2 + (y-0)^2 = (x+6)^2 + (y+3)^2
$$
Multiplying out, we get
$$
x^2 -8x +16 + y^2 = x^2 +12x +36 \; + \; y^2 +6y + 9
$$
So
$$
20x + 6y + 29 = 0
$$
