What You Need to Know About a Straight Line
A.
Necessary Concepts, Forms of the Equation, and Formulas
The slope of a line is the tilt or slant.
slope  m 
rise change in y y y 2  y1



run change in x x x2  x1
For the points
(x1, y1) and
(x2, y2)
Ax  By  C
Standard form for the equation of a line is:
Notice this is just the x term first, then the y term, then the equals sign, and finally the constant term.
For Example: 3x  5 y  7
The equation of a line may also be written in slope-intercept form:
y  mx  b
Notice this equation is solved for y.
For Example: y  2 x  4
Lines with a positive slope rise from lower left to upper right.
Lines with a negative slope fall from upper left to lower right.
Two types of lines are special:
Vertical Lines have the form x
= a and have no slope or undefined slope.
For Example: x  5 will be vertical and have no slope.
Horizontal Lines have the form y
a and b are
real numbers.
= b and a slope of zero.
For Example: y  3 will be horizontal and have a slope of zero.
The point-slope form of the equation is most often used to find the equation of a given line:
y  y1  mx  x1  for point (x1, y1) and slope of m.
Parallel lines have the same slope (they are slanted the same).
Perpendicular lines have negative reciprocal slopes. This means the slopes must meet two
conditions: opposite in sign ( +/-) and “flipped over”/reciprocals.
B.
Types of Problems Concerning Straight Lines
1. Find the slope of a line:
a. Given Two Points (x1, y1) and (x2, y2), use the formula
For Example: ( 2, -3) and ( -4, 7)
What You Need to Know About a Straight Line
b. Given the graph: Use the formula
For Example:
Locate 2
definite
points on the
graph.
Rise is Up (+) or Down (-).
Run is Left (-) or Right (+).
Note the directions you
must travel to move from
one point to the other.
Starting with the lower left
point we would travel UP 1
and RIGHT 2. Both are +.
Rise in the numerator and
run in the denominator:
c. Given the equation in slope-intercept form: Use y = mx + b knowing m = slope
For Example: y = 3x – 7
The slope is 3 since in m = 3.
d. Given the equation in standard form (Ax + By = C): Use
For Example: 4x - 6y = 7,
The slope will be
2. Find the equation of a line:
a. Given a point and the slope:
Use the point-slope form: y  y1  mx  x1 
For Example: m = -3 and the line passes through (-4,5)
Substituting into the point-slope form gives us:
y - 5 = -3(x - (-4))
y - 5 = -3(x + 4)
y - 5 = -3x - 12
In standard form, we would have 3x + y = -7
In slope-intercept form, we would have y = -3x - 7
b. Given two points :
Use the formula
to find the slope. Then use this slope with either of
the two points in point-slope form, y  y1  mx  x1  , as shown above in part a.
c. Given the slope and the y-intercept:
Simply substitute both into the slope-intercept form of the equation.
For Example:
and the y-intercept is -4
Since
and b = - 4, we may substitute into y = mx + b
What You Need to Know About a Straight Line
d. Given x- and y-intercepts:
This is the same as being given 2 points. If the x-intercept is -3 and the yintercept is 7, remember the actual coordinates of these points are then (-3,0) and
(0,7). These are the coordinates for two points, so work the problem as in part b
above.
e. Given a point and a line parallel or perpendicular:
Find the slope of the given line. If parallel, then use this slope. If perpendicular,
use the opposite reciprocal. Then use point-slope form y  y1  mx  x1  with
the point and slope.
For Example: Line passing through ( 2, -1 ) and is parallel to y = 4x – 8
Since the lines are parallel, they have the same slope so m = 4. Now, we
have a point ( 2, -1) and slope so use y  y1  mx  x1  .
y – ( -1) = 4( x – 2)
y + 1 = 4x – 8
y = 4x – 9
For Example: Line passing through ( 2, -1 ) and is perpendicular to y = 4x – 8
Since the lines are perpendicular, they have opposite reciprocal slopes.
Therefore, since m = 4 of the given line, the slope needed is
.
Now, we have a point ( 2, -1) and slope so use y  y1  mx  x1  .
y – ( -1) =
( x – 2)
y +1=
y=
3. Graph the line:
a. Given Slope-Intercept Form, y = mx + b:
Follow these steps:
1. Plot the point which is the y-intercept on the y – axis.
2. Use the slope as directions to a new point using
3. Draw the line connecting the points.
.
For Example: Graph the line y = -3x + 2
The slope, m, is . We need it as a fraction to have a
rise and a run. The y-intercept, b, is 2 so (0, 2) is a
point on our graph.
*See step images on the next page.
What You Need to Know About a Straight Line
Example continued:
1. Plot b on the y – axis (0, 2).
2. Use
=
.
3. Draw the line.
Down 3
and
Right 1
b. Given Standard Form, Ax + By = C:
Standard form can be converted to slope-intercept form by solving the equation
for y. Then, the previous method would be utilized to graph the line. Otherwise,
follow these steps to graph the line from standard form.
Follow these steps:
1. Determine the y-intercept, ( 0, y), of the line by plugging in 0 for x in the
equation and then solving for y.
2. Determine the x-intercept, ( x, 0), of the line by plugging in 0 for y in the
equation and then solving for x.
3. Plot these two points on the graph.
4. Draw the line connecting the points.
For Example: Graph the line 2x + 4y = 12
Plug 0 in for x to get 2(0) + 4y = 12. Now, solve for y. y = 3.
Therefore, the y-intercept is (0, 3). Next, plug 0 in for y to obtain 2x +
4(0) = 12. By solving for x, the result is x = 6. Thus the x-intercept is
(6, 0). Plot both points on the graph and draw the line.
1.
Plot the y-intercept (0, 3).
2. Plot the x-intercept (6, 0).
3. Draw the line.