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?

Notice, the mid=point of the line joining $(4, 0)$ & $(-6, -3)$ is given as  $$\left(\frac{4+(-6)}{2}, \frac{0+(-3)}{2}\right)\equiv \left(-1, -\frac{3}{2}\right)$$ The slope of the perpendicular bisector 
$$=\frac{-1}{\text{slope of line joining}\ (4, 0)\ \text{&}\ (-6, -3)}$$ 
$$=\frac{-1}{\frac{-3-0}{-6-4}}=-\frac{10}{3}$$
Hence, the equation of the perpendicular bisector: $$y-\left(-\frac{3}{2}\right)=-\frac{10}{3}(x-(-1))$$  $$6y+9=-20x-20$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{equation of the perpendicular bisector:}\ 20x+6y+29=0}}$$
