# SOLUTION: I need to know if I have done these right!! Thank you so much!! ) Given a line containing the points (1,3), (2,4), (3,5) determine the slope-intercept form of the equation, giv

Algebra ->  Graphs -> SOLUTION: I need to know if I have done these right!! Thank you so much!! ) Given a line containing the points (1,3), (2,4), (3,5) determine the slope-intercept form of the equation, giv      Log On

Question 73680This question is from textbook college algebra
: I need to know if I have done these right!! Thank you so much!!
) Given a line containing the points (1,3), (2,4), (3,5) determine the slope-intercept form of the equation, give one additional point on this line and graph the function.

Show your work here: y2-y1/x2-x1= -4,-6

Give one additional point in (x,y) form that would fall on this line: 4,6

This question is from textbook college algebra

Found 2 solutions by jim_thompson5910, bucky:
You can put this solution on YOUR website!
 Solved by pluggable solver: Finding the Equation of a Line First lets find the slope through the points (,) and (,) Start with the slope formula (note: (,) is the first point (,) and (,) is the second point (,)) Plug in ,,, (these are the coordinates of given points) Subtract the terms in the numerator to get . Subtract the terms in the denominator to get So the slope is ------------------------------------------------ Now let's use the point-slope formula to find the equation of the line: ------Point-Slope Formula------ where is the slope, and (,) is one of the given points So lets use the Point-Slope Formula to find the equation of the line Plug in , , and (these values are given) Distribute Multiply and to get . Now reduce to get Add to both sides to isolate y Combine like terms and to get ------------------------------------------------------------------------------------------------------------ Answer: So the equation of the line which goes through the points (,) and (,) is: The equation is now in form (which is slope-intercept form) where the slope is and the y-intercept is Notice if we graph the equation and plot the points (,) and (,), we get this: (note: if you need help with graphing, check out this solver) Graph of through the points (,) and (,) Notice how the two points lie on the line. This graphically verifies our answer.

So we know the equation is . Let's plug in another x value. If we let x=4 then:

So another point on this line is (4,6)
So yes you did this right. You just need to find the equation and graph it.

You can put this solution on YOUR website!
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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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 and another of the points as .
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It's a little easier to see if we choose as a point that is to
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the right of the point we call . 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 the point (3,5). By comparison, that means
and .
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Now let's choose as the point (1,3). By comparison of these two
that means and .
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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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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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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 . 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 . Solve this by subtracting 1 from both
sides to get:
.

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Then take this value for b and plug it into the slope intercept form of and
you have the final version of the slope intercept form as being:
.

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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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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.