Question 771401
Using some general point on a line, (u,v), the the point-slope form of a line would be {{{y-v=m(x-u)}}}, from which you can make {{{y=m(x-u)+v}}}


Using that general point-slope formula for a line, you find line A this way:
{{{y=-4(x-(-25))+(-10)}}}
{{{y=-4x-100-10}}}
{{{highlight(y=-4x-110)}}} ________________Line A.


You can do similarly for line B.


Question (a):  How far is line A to (8, -3)?
The best way to start this one is, find the line containing (8, -3) which is perpendicular to  line A.  In finding this, initially we do not know the y-intercept.
We want slope {{{(1/4)}}}.
{{{y=(1/4)x+b}}} where unknown y-intercept is b.
{{{b=y-(1/4)x}}}
{{{b=(-3)-(1/4)*8}}}
{{{b=-5}}}
This perpendicular line to A is {{{highlight(y=(1/4)x-5)}}}.


What is intersection point of {{{y=(1/4)x-5}}} and {{{y=-4x-110}}} ?
Equating the y formulas, solving for x, then returning and solving for y,
these line intersect at.... (-420/17, 20/17).
Use distance formula to find the distance from (-420/17, 20/17) to (8, -3).


{{{highlight(D=sqrt((-420/17-8)^2+(20/17-(-3))^2))}}}
Remaining calculation not shown...)