Question 34632
There is a lot to this question, so lets first start with parts a and b.  I am not sure if you have to have a specific or general equation.  Standard form of an equation is ax + by = c.  You can have any numbers for a,b,and c.  Slope intercept form is y=mx+b and m and b could be any numbers.  For #38- it wants you to find an equation that goes through the point (-2,0) and is perpendicular to 8x-3y=7.  I hope that makes sense- maybe it will make more as we keep going.  Lets start with 8x-3y=7.  In order for two lines to be perpendicular, the product of their slopes has to be -1, or they have to be inverses.  Lets first find the slope of 8x-3y=7.  Put that equation in slope intercept form: 
8x - 3y = 7
-3y = -8x + 7
y = 8/3x - 7/3
The only important number in this equation is 8/3 - that is the slope of this equation.  So, the slope of a line perpendicular to that would be its inverse or -3/8.  

Now, we have the slope of the line and a point that this line goes through (-2,0).   Lets use point slope form (because we know a point and the slope) to find the equation of the line.  POint slope is {{{m=(y-y1)/(x-x1)}}}where (x1,y1) is the point on our line. 
Lets substitute: {{{(-3/8)=(y - 0)/(x - (-2))}}}
Now solve for y to get it into slope intercept:
{{{(-3/8)=(y - 0)/(x - (-2))}}}
{{{(-3/8)=(y)/(x + 2))}}}
{{{(-3/8)*(x + 2)=y}}}
{{{(-3/8)*x + (-3/8)*2 = y}}}
{{{(-3/8)*x + (-3/4) = y}}}
Now, to get in standard form, put x back on the same side with y and get rid of your fraction:
{{{(-3/8)*x + (-3/4) = y}}}
{{{(-3/4) = y + (3/8)*x}}}
Multiply everything by 8 to get rid of fractions
{{{8*(-3/4) =8y + 8(3/8)*x}}}
{{{-6= 8y + 3x}}}