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

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:
Answer by jim_thompson5910(35256) (Show Source): 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.
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
I'm sorry it took so long for me to get to your problem.
.
The slope-intercept form of an equation is:
.
y = mx + b
.
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.
.
So let's find the slope. You have the equation correct for finding the slope. That equation is:
.

.
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 .
.
It's a little easier to see if we choose as a point that is to
.
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.
.
Anyhow let's choose as the point (3,5). By comparison, that means
and .
.
Now let's choose as the point (1,3). By comparison of these two
that means and .
.
Then all we do to calculate the slope is to substitute these 4 values into the slope equation
as follows:
.

.
So we have the slope of m = 1. Plug that value of m into the slope intercept form of the
equation and you get:
.
.
.
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:
.

.
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:
.

.
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:
.

.
Well, that certainly is true, so your point of (4,6) IS on the graph.
.
Hope this helps you to understand the first part of the problem a little better.
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