Help with multiple questions

1. Line $l$ has equation $y = 4x - 7$, and line $m$ with equation $y = ax + b$ is perpendicular to line $l$ at $(2,1)$. What is the $y$-coordinate of the point on $m$ that has $x$-coordinate 6?

The slope of the perpedicular line  is    - (1/4)

So...the equation of  line   m  is

y  = (-1/4)(x -2)  + 1

y  = (-1/4)x + 1/2 + 1

y  = (-1/4)x + 3/2 

So  on this line....when x  = 6

y  = (-1/4)(6) + 3/2

y  = -6/4 + 3/0

y = -3/2  + 3/2

y = 0

2. Find $B - A$ if the graph of $Ax + By = 7$ passes through $(2,1)$ and is parallel to the graph of $2x - 7y = 3$.

The slope of the first line  =  -A/B  =  the slope of the second line  = 2/7

Which implies that   -7A  = 2B  ⇒  B  = (-7/2)A

So

A(2)  + B(1)  = 7

A(2)   +  (-7/2)A  = 7

-(3/2)A  =  7

A  =  -14/3

B =  (-7/2)(-14/3)  =  98/6  =  49/3

So.......the equation of the first line is

(-14/3)x  + (49/3)y  = 7

And B - A  =     49/3 + 14/3      =   63 / 3  =  21

Here's the graph showing this : https://www.desmos.com/calculator/qrhzf2f3op

3.The "perpendicular bisector" of the line segment $\overline{AB}$ is the line that passes through the midpoint of $\overline{AB}$ and is perpendicular to $\overline{AB}$.  The equation of the perpendicular bisector of the line segment joining the points $(1,2)$ and $(-5,12)$ is $y = mx + b$. Find m + b

The midpoint of AB  is  [ (1 - 5) / 2, (12 + 2) / 2    =  ( - 4/2 14/2)  =  (-2, 7)

And the slope of AB  =  [ 12 - 2] / [ -5 - 1 ]   =  10 / -6   =   - 5/3

So....the perpedicular bisector will have the slope  3/5

And the equation of this bisector  is

y  = (3/5)(x - - 2) + 7

y  = (3/5)x + 6/5 + 7

 y = (3/5)x + 41/5

So.....m + b  =    3/5  + 41/5  =   44/5