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?
 A: lol. a lot of confusion on this thread. When you are computing the midpoints, what you actually do is find the distance between them and divide by two. This isn't the coordinate of the midpoint; this is just the distance form one end to the midpoint.
What you meant to do is take the average of the x and y values, not the average of their difference.
You should get a midpoint of $(-1,-1.5)$.
A: To find the midpoint, you don't need to negate the coordinates. So the midpoint is $\left(\frac{-6+4}{2}, \frac{0+(-3)}{2}\right)=\left(-1, \frac{-3}{2}\right)$
A: First you computed midpoint of $X$ wrongly.
Later you evaluated $b$ wrongly.
A: 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: $$y-\left(-\frac{3}{2}\right)=-\frac{10}{3}(x-(-1))$$  $$6y+9=-20x-20$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{equation of the perpendicular bisector:}\ 20x+6y+29=0}}$$
A: 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
$$
20x + 6y + 29 = 0
$$
