Question 73680
I'm sorry it took so long for me to get to your problem.
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The slope-intercept form of an equation is:
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y = mx + b
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where m, the multiplier of x, is the slope of the graph, and b is the point on the y-axis where
the graph crosses.
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So let's find the slope. You have the equation correct for finding the slope. That equation is:
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{{{m = (y[2] - y[1])/(x[2]-x[1])}}}
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We know that all three of the given points are on the line.  All we have to do is identify
one of the points as {{{(x[2]y[2])}}} and another of the points as {{{(x[1] y[1])}}}.
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It's a little easier to see if we choose as {{{(x[2] y[2])}}} a point that is to
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the right of the point we call {{{(x[1]y[1])}}}. It doesn't have to be that way because it 
will work out the same no matter what.
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Anyhow let's choose as {{{(x[2] y[2])}}} the point  (3,5). By comparison, that means
{{{x[2] = 3}}} and {{{y[2] = 5}}}.
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Now let's choose as {{{(x[1] y[1])}}} the point (1,3).  By comparison of these two 
that means {{{x[1] = 1}}} and {{{y[1] = 3}}}. 
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Then all we do to calculate the slope is to substitute these 4 values into the slope equation
as follows:
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{{{m = (y[2] - y[1])/(x[2]-x[1]) = (5 - 3)/(3 - 1) = (2/2) = 1)}}}
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So we have the slope of m = 1.  Plug that value of m into the slope intercept form of the
equation and you get:
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{{{ y = mx+ b =1*x +b = x + b}}}.
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To solve for b, all we need to do is to take one of the given points and plug its x and y
values into the equation {{{ y = x + b}}}.  Then we can solve for b.  For example,
let's take the point (1,3) that we were given.  Plug 1 in for x and 3 in for y and the
slope intercept equation becomes {{{3 = 1 + b}}}. Solve this by subtracting 1 from both
sides to get:
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{{{2 = b}}}
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Then take this value for b and plug it into the slope intercept form of {{{y = x + b}}} and
you have the final version of the slope intercept form as being:
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{{{y = x + 2}}}
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Next you said that you thought the point (4,6) was on the graph. Let's check that out by
putting 4 into our slope intercept equation for x and 6 in for y and see if the equation
is still true.  When we do the substitution we get:
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{{{6 = 4 + 2}}}
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Well, that certainly is true, so your point of (4,6) IS on the graph.
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Hope this helps you to understand the first part of the problem a little better.