<PRE><B>Question 21909</B>
I don&#39;t really understand what section you are talking about.
But I will give you some tips, you can email me with examples of what you are having trouble with and I will try to help. athompson022@ntcc.edu.
First, for slope you take the two sets of points you are given and put them like this and solve for slope
slope = y2 - y1       your points will be shown as (x-y) (x2-y2) i only use 2
        -------       to indicate which point you put in the first set of point
        X2 - X1       are x and y, the second set is x2 and y2, but do not use 
                      the 1 or 2s in your problem.  for example points (1,1)   
                      and  (2,4) you would set this up to find slope as follows
m= 4-1
   ---- = 3/1=3  so m=3   3 is your slope
   2-1
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To find the y intercept for an equation set x in your equation = 0 then solve for y. your answer would be something like (0,y). to find the x intercept set y=0 in your equation and slolve for x your answer would be (x,0).  get it?
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an example of a question that was on my test for some of the things you may need are given below.  I don&#39;t really know what you need, so maybe these will help.

1.  given -3x + 3y = 12, find the slope and y intercept.
-3x+3y=12    you want this in the form y=mx+b  with m being your slope
3y=3x+12
y=3x/3 + 12/3   to get rid of the 3 in front of the y
y=x+4
your slope would be 1 since there is only 1x.    
                to get the yintercept, set x equal to zero and solve for y.
                the answer would be (0,y) this is the first point to plot on 
                graph. since 3 is your slope you use 3/1. you have plotted 
                your first point you move up 3 and over 1 to get the next point
                on the graph. then repeat the up 3 and over 1 for all the 
                points you need.
0 + 3y=12
3y=12
y=4  your first point would be (0,4), your y intercept
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2.  you are asked to find the equation of the line passing through (6,-3) and parallel to 2y-6x=12
Remember if they are parallel they have the same slope.  First put the equation into y=mx+b form
2y=6x+12    you must get rid of the 2 in front of the y by dividing both sides
            by 2
y=3x+4   this is in y=mx+b form the slope is the m value.(#in front of x)
            you now have points and a slope and can use point-slope formula
            y-y2=m(x-x1)
get the y1 and x1 values from your given points (6,-3)

y+(-3)=3(x-6)      this is in point slope formula
y+3=3x-18
y=3x-3-18
y=3x-21     this is the equation of a line (y=mx+b) notice that the slopes are the same because they are parallel lines.
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For a perpendicular line, the slope is always the negative reciprocal of the slope of the first line.
for example
3.  find the equation of the line passing through(5,3) and perpendicular to x=5y-7.  
     first you must set the equation in y=mx+b form
-5y=-x-7    I find it easier to change all the signs
5y=x+7      you actually divide each side by -1
y=1/5x+7    this is in y=mx+b 
            because the line is perpendicular, you use the negative reciprocal of 1/5 as your slope.  that is -5
put it in the slope intercept form because you have a slope now and a set of points.
y-3=-5(x-5)
y-3=-5x+25
y=-5x+25+5
y=-5x+30     this is your equation for a perpendicular line in y=mx+b



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