# Straight Lines_1_ by yaofenjin

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```									                              Straight Lines

To graph a straight line:
The equation of a straight line is y = mx+b; b indicates where the graph crosses the y-axis
and m is called the slope, or rise over run ratio. y = mx+b is called the slope-intercept
equation and is the most important thing you have to know about straight lines. To see
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how to graph a straight line like y  x  2 , go through the following steps:
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1)     The 2 indicates that the line crosses the y-axis at the point (0,2). This point is
called the y-intercept.

2)     From the y-intercept, count 5 steps over to the right. 5 is the run of this line and
it is best counted from left to right.

3)     From where you are now, 5 steps to the right of the y-intercept, count 3 steps up.
Mark the point where you are now and draw the line passing through the y-
intercept and the point you just marked. After you have done this accurately on
graph paper your graph will look like this:

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Should the slope, i.e. the number in front of the x, be negative, then count the rise
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downwards. Thus, y   x  1 is graphed by starting from a y-intercept of (0,1),
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counting 7 steps to the right and then 2 steps down. This line will look like this:

A line given as 4x+3y = -9 which is not in the slope intercept form, can best be plotted by
first solving this equation for y and then going through the steps above. The line could be
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written in the slope intercept form as y   x  3 .
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A few details deserve to be mentioned here:

1)     Vertical lines have no slope and have an equation like x = a with (a,0) as the x-
intercept.

2)     Horizontal lines have a slope equal zero and an equation like y = b with (0,b) as
the y-intercept.

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3)     In case of a line with integer slope like y = 4x - 3, write it as y      x  3 and then
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graph it using the above outlined procedure.

4)     In case of a line with a decimal slope, first change the decimal to a fraction. So
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the line y = 0.6x - 2 should first be written as y  x  2 .
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Problem type: How do I evaluate the x- and the y-
intercepts of a line?

You can always find the x-intercept by first setting y = 0 and then solving for x. So to
find the x-intercept of 4y – 3x = 8, set y = 0 so that -3x = 8 and you find an x-intercept of
(-8/3,0).

You can always find the y-intercept of a line by first setting x = 0 and then solving for y.
Thus, the y-intercept of 4y – 3x = 8 is found from 4y = 8 so that (0,2) is the y-intercept.
Of course you could also have put this line in slope intercept form to find the y-
intercept.

The procedures for finding the intercepts do not only hold for straight lines but also for
any other sort of graph.

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Problem type: How do I find the slope of a straight line
given that this line passes through two
points?
This will be easy after making a graph. Take a piece of graph paper, draw a coordinate
system, plot the two given points and then count the run from the left most point to the
right most and lastly count the rise. If you're counting the rise upwards then the slope is
positive. Should you have to count the rise downwards then the slope is negative.
Remember? The slope is the rise over run ratio of the line. For this to work you should
always count the run first, from left to right.

Example: Given that a straight line passes through (-5,4) and (2,-1), what is the slope of
this line? Look at the graph below. Using graph paper you can easily count the
run as 7 and the rise as 5 (or -5). Thus, the slope is -5/7.

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Problem type: How do I find the equation of a straight
line given the slope and one point on the
line?

Easy, start from the slope intercept equation y = mx + b and substitute m, x and y. Then
solve for b.

Example: Find the equation of the line with slope 5/2 and containing the point (-3,4).

Start from the slope intercept form and substitute y = 4, m = 5/2 and x = -3.
This gives us: 4    3  b so that 4    b and b 
5                         15              23
. The equation
2                          2               2
5     23
we are looking for is y  x 
2     2

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Problem type: How do I find the equation of a straight
line given two points on the line?

Do this in two steps. First find the slope of the line using a sketch and counting rise and
run. Next, start from the slope intercept form, substitute m, and the x and y from one of
the given points, and then find b. Which of the two points you choose to use does not
matter. This will give you the equation of the line.

Example: Find the equation of the straight line containing the points (-3,-2) and (5,-4).

First plot the two points on graph paper and draw the line through them. Next,
starting from (-3,-2) count the run and rise. As you can see from the graph
below, this gives us m=-2/8=-1/4.

Next, choose the point you would like to use in your substitution. Let us select (-3,-2).
Now substitute the slope and the x and y values of this point in to the slope intercept form
to get  2     3  b so that  2   b and thus b   . So the equation of the
1                           3                  11
4                           4                   4
1     11
line we are looking for is y   x  .
4      4
As an afterthought consider the following. What is the use of making a graph when you
can easily find the slope without it - see your textbook on how to do just that. But
making the graph will serve yet another purpose. Look at the graph and at the y-intercept.
Now check your calculations. What is the y-intercept? As you see, the graph will help
you to check your work. If you mess up in your calculations in finding the slope or the y-
intercept, or even both, this will show when you compare your graph to your calculations.
Keep this in mind.

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Problem type: How do I find the equation of a straight
line given that the line contains a certain
point and that it is parallel with respect
to a given line?

Whenever two lines are parallel, they will have equal slopes. Start from y = mx + b,
substitute the slope of the given line and solve for b in the usual way.

Example: Find the equation of the line containing (2,-3) and parallel with respect to 4x-
3y=12.

First, solve the given line for y so you know its slope and thus the slope of the
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line you're looking for. This gives us y  x  4 . Use the slope of this line
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and the x and y from the given point, and then find b. This gives us
4                     17                                     4    17
 3   2  b so that b   . So the equation of the line is y  x 
3                      3                                     3     3

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Problem type: How do I find the equation of a straight
line given that the line contains a certain
point and that it is perpendicular with
respect to a certain line?

To solve this sort of problem you need to know that when two lines are perpendicular
with respect to each other, the product of their slopes should equal -1. So the slope of
4                   3
any line perpendicular with respect to the line y  x  2 must have  as its slope,
3                   4
4  3
since      1 .
3  4

Example: Find the equation of the straight line containing (-2,1) and perpendicular to the
line 3y+2x=6.

First, write the given line in slope intercept form to see what its slope is. You
2                                                          3
will get y   x  2 . So the slope of any perpendicular line must be . Now
3                                                          2
find b using the now familiar technique, and get 4. The line we are looking for
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has an equation y  x  4 .
2

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