We are often given the problem of finding the equation of a line given two distinct points on the line.
The solution is almost a mechanical process. The following flowchart illustrates the steps:
Figure 1
| We can start by pressing the keys
to open the VAR-LINK menu.
The calculator used to generate Figure 1 holds a large
number of programs. Within folders the program and variable names
are displayed in alphabetic order.
We are looking for the "slope" program, and we can have the display shift to the
next item that starts with the letter "s" by pressing the
key. In fact, that key was pressed twice to arrive at
the image shown in Figure 1.
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Figure 2
| Figure 2 reflects the condition of having found the
slope program in the VAR-LINK window, and of pressing the
key to select that program and to paste its name into the
command entry and edit line of the HOME screen. Actually, what gets pasted there is "slope(".
We need to complete the command, to enter a right parenthesis, by
pressing the key.
We can execute the command and start running the program by
pressing the key.
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Figure 3
| In Figure 3, the program has started.
Output from the program appears in the PROGRAM IO window.
The program clears the screen and
displays "Get Slope, etc." on the top line, followed by a
prompt for the value of "x1" on the second line.
We are looking for the solution to "Find the equation of the line containing the points
(4,7) and (10,1)." Therefore, we respond with
for x1.
The program accepts that value and continues by asking for the value
for y1.
assigns 7 to y1.
This is followed by
to leave the screen as it appears
in Figure 3. At this point we are in danger of losing the top portion of the screen
if we go any further.
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Figure 4
| We press
to give x2 the value 10. Then we press
to create the value for y2.
To accpet that value and move to Figure 5 we press the
key.
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Figure 5
| The slope program responds by repeating the input points,
computing and displaying the slope of the line (in this case, 1),
computing and displaying both the slope-intercept form and the standard form of the
desired equation.
Having produced all of the answers, the program is in a "paused" condition, as indicated by
the displayed in the lower right corner of
the screen. We can leave this screen and continue with the next set of
input points by pressing the key.
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Figure 6
| Rather than present screen images that show the data input, we will
skip to the main output screen. In this case we are solving the problem:
"Find the equation of the line containing the points
(5,1) and (8,10)." Once the values have been given, we press
to accept the final value and move to Figure 6.
The calculator has determined the slope, the slope-intercept form
of the solution, and the standard form of the solution.
Please note that the slope program produces a standard form version of
the equation that does not require the leading coefficient to be non-negative.
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Figure 7
| Figure 7 follows from the data input for yet another example:
"Find the equation of the line containing the points
(2,7) and (5,9)."
The calculator has determined the slope between the two points to be
2/3. Note that the calculator program has taken the liberty of
writng the slope-intercept form as y=2/3 x + 17/3.
It would have been better to have included parentheses so that the answer appears as
y=(2/3) x + 17/3, leaving no doubt but that the "x"
is not in the denominator of the fraction. Once we see and understand the somewhat "sloppy"
form that the program uses for fractional slopes, we can correctly understand the
equations that the program produces.
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Figure 8
| Figure 8 follows from the data input for yet another example:
"Find the equation of the line containing the points
( 4,3) and ( 7,25)."
The calculator has determined the slope between the two points to be
22/3. Note that
in the slope-intercept form of the solution equation, the calculator program has
left the slope and the second coordinate of the y-intercept as improper fractions.
slope does not convert these to either mixed numbers or to decimal values.
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Figure 9
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Figure 9 represents the data input for yet another example:
"Find the equation of the line containing the points
(2,9) and (8,9)." By inspection we can see
that the solution will be a horizontal line. The slope
program correctly identifies the equation as the constant function,
with the solution equation being y=9.
Note that we could write this as y=0x+9 to produce the strict slope intercept form.
In addition we could write this as 0x + 1y = 9 to produce the strict standard
form of the equation.
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Figure 10
| Figure 13 represents the data input for yet another example:
"Find the equation of the line containing the points
(6,1) and (6,10)." By inspection we can see
that the solution will be a vertical line.
The slope program correctly identifies the equation as a vertical line
with the solution equation being x=6.
Note that we could write this as 1x + 0y = 6 to produce the strict standard
form of the equation.
It is impossible to construct a slope-intecept form of this equation: the slope is undefined and the
coefficient of y is 0.
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Figure 11
| If we want to stop the program we need to break out of it. We can press the
key to break out of the program. The result is shown in Figure 11.
There are only two options. Either we can GOTO or QUIT. We want to
QUIT, so we press the key to move to Figure 12.
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Figure 12
| Figure 12 looks remarkably similar to Figure 10.
However, in Figure 10 the calculator
was in a "paused" condition.
In Figure 12 the program has terminated. However, we are still in the
Program IO screen. We will need to press the
key to move back to the HOME screen, shown in Figure 13.
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Figure 13
| Figure 13 shows the updated HOME screen. Note tht we have a record of having run the slope
program and of having terminated that program through a "break".
In addition, the previous command, slope(), remains in the command input and edit line. |
Figure 14
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Figure 14 shows that the program has been restarted.
Having broken out of the program in Figures 11 through 13, we only needed to
press the key to re-issue the
the last command, namely slope(), and therefore restart the program.
In Figure 14 we have entered the two points derived from the discussion above.
The first thing to notice in Figure 14 is that the slope has been computed to be
exactly the value given in the problem statement. This is verification that we have correctly computed the
coordinates of the second point. Then we can look at the slope-intercept and
standard forms for the desired equation.
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Figure 15
| The next problem to consider is
Find the equation of the line that has slope=2 and that contains the point ( 4,1).
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In order to use the slope program, we need to know two points on the line.
The problem statement gives us one point, ( 4,1). Can we get a second point?
Yes. We know the slope and we know that
m = |
2 | = | change in y |
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|
1 | change in x |
Therefore, if we start at the point ( 4,1)
and we add 1 to the x value and add 2 to the y value
we produce a new point ( 3,3) that must be on the desired line.
Thus, we use the definition of slope to produce
a second point, and this allows us to use the slope program.
The values for the two points
have been entered into the program to arrive at Figure 15.
The screen
confirms the second point by calculating the expected slope. Then, the slope
program produces the slope-intercept and the standard forms of the desired equations.
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Figure 16
| The next problem to consider is
Find the equation of the line that has slope= 3/5
and that contains the point ( 6, 13).
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In order to use the slope program, we need to know two points on the line.
The problem statement gives us one point, ( 6, 13).
Can we get a second point?
Yes. We know the slope and we know that
m = |
3 | = | change in y |
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|
5 | change in x |
Therefore, if we start at the point ( 6, 13) and
we add 5 to the x value and add 3 to the y value
we produce a new point ( 1, 16) that must be on the desired line.
Thus, we use the definition of slope to produce
a second point, and this allows us to use the slope program. The values for the two points
have been entered into the program in Figure 16. In fact, Figure 16 shows the state of the screen
after we have entered the value for y2, but before we have accpeted that value.
We can observe the coordinates of the second point in Figure 16.
We press the key to accept that value and move to Figure 17.
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Figure 17
| Figure 20 confirms the second point by calculating the expected slope. Then, the SLOPE
program produces the slope-intercept and the standard forms of the desired equations.
Notice that the coordinates of the second point have been cut off on the right side of the screen.
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