Example 1 Determine the slope of each of the following lines. Sketch the
graph of each line.
.
(a)
The line that contains the two points (−2,−3) and (3,1) .
(b)
The line that contains the two points (−1,5) and
(0,−2) .
(c)
The line that contains the two points (−3, 2) and
(5, 2) .
Solution
Okay, for each of these all that we’ll need to do is
use the slope formula to find the slope and then plot the two points and connect
them with a line to get the graph.
(a) The line that contains
the two points (−2,−3) and
(3,1) .
Do not worry which point gets the subscript of 1 and
which gets the subscript of 2. Either way will
get the same answer. Typically, we’ll just take them in the order listed. So,
here is the slope for this part.
Be careful with minus signs in these computations. It is easy to
lose track of them. Also, when the slope is a fraction, as it is here, leave it as a fraction. Do
not convert to a decimal unless you absolutely have to.
Here is a sketch of the line.
Notice that this line increases as we move from left to
right.
(b) The line that contains
the two points (−1,5) and (0,−2) .
Here is the slope for this part.
Again, watch out
for minus signs. Here is a sketch of the graph.
This
line decreases as we move from left to right.
(c)
The line that contains the two points (−3, 2) and (5, 2) .
Here is the slope for this line.
We got a slope of zero here. That is okay, it will happen
sometimes. Here is the sketch of the line.
In this case we’ve got
a horizontal line.
(d)
The line that contains the two points (4,3) and (4,−2) .
The final part. Here is the slope computation.
In this case we get division by zero which is undefined. Again,
don’t worry too much about this it will happen on occasion. Here is a sketch of
this line.
This final line is a vertical line.
Example 2 Write down the
equation of the line that passes through the two points (−2, 4) and (3,−5) .
Solution
At first glance it may not appear that we’ll be able to use the point-slope
form of the line since this requires a single point (we’ve got two) and the slope (which
we don’t have). However, that fact that we’ve got two points isn’t really a problem; in fact, we
can use these two points to determine the missing slope of the line since we do
know that we can always find that from any two points on the line.
So, let’s start off my finding the slope of the line.
Now, which point should we use to write down the equation of the
line? We can actually use either point. To show this we will use both.
First, we’ll use (−2, 4) .
Now that we’ve gotten the point all that we need to do is plug into the formula. We will use the second form.
Now, let’s use (3,−5) .
Okay, we claimed that it wouldn’t matter which point we used in the
formula, but these sure look like different equations. It turns out however,
that these really are the same equation. To see this all that we need to do is
distribute the slope through the parenthesis and then simplify.
Here is the first equation.
Here is the second equation.
So, sure enough they are the same equation.
Example 3 Determine the slope of each of the following equations and sketch
the graph of the
line.
(a)
2y − 6x
= −2
(b)
3y + 4x
= 6
Solution
Okay, to get the slope we’ll first put each of these in
slope-intercept form and then the slope will simply be the coefficient of the x
(including sign). To graph the line we know the y-intercept of the
line, that’s the number without an x (including sign) and as discussed
above we can use the slope to find a second point on the line. At that point
there isn’t anything to do other than sketch the line.
(a)
2y − 6x
= −2
First solve the equation for y. Remember that we
solved equations like this back
in the previous
chapter.
2y = 6x – 2
y = 3x – 1
So, the slope for this line is 3 and the y-intercept
is the point (0,−1) .
Don’t forget to take the sign when determining the y-intercept. Now, to
find the second point we usually like the slope written as a fraction to make it clear what the rise
and run are. So,
The second point is then,
x2 = 0 + 1 = 1 y
= −1 + 3 = 2 ⇒ (1,2)
Here is a sketch of the graph of the line.
(b)
3y
+ 4x
= 6
Again, solve for y.
In this case the slope is – 4/3 and the y-intercept is (0, 2) . As
with the previous part let’s first determine the rise and the run.
Again, remember that if the slope is negative make sure that the
minus sign goes with the numerator. The second point is then,
x2 = 0 + 3 = 3 y
= 2 + (– 4) = – 2 ⇒ (3, – 2)
Here is the sketch of the graph for this line.
Example 4 Determine the equation of the line that passes through the point (8, 2) and
is,
(a)
Parallel to the line given by 10y + 3x = −2
(b)
Perpendicular to the line given by 10y + 3x = −2 .
Solution
In both parts we are going to need the slope of the
line given by 10y + 3x
= −2 so let’s actually find that before we get into the individual parts.
10y = – 3x – 2
Now, let’s work the example.
(a)
parallel to the line given by 10y
+ 3x
= −2
In this case the new line is to be parallel to the line given by 10y
+ 3x
= −2
and so it must have the same slope as this line. Therefore we know that,
Now, we’ve got a point on the new line, (8, 2),
and we know the slope of the new line, –
3/10, so we can use the point-slope form of the line to write down the equation
of the new line.
For this part we want the line to be perpendicular to 10y + 3x
= −2
and so we know we can find the new slope as follows,
Then, just as we did in the previous part we can use the
point-slope form of the line to get the equation of the new line. Here it is,
.
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