M098
Carson Elementary and Intermediate Algebra 3e
Section 4.5
Objectives
1.
2.
3.
Use point-slope form to write the equation of a line.
Write the equation of a line parallel to a given line.
Write the equation of a line perpendicular to a given line.
Vocabulary
Parallel lines
Perpendicular
lines
Lines in the same plane that never intersect. They have equal slopes.
Lines in the same plane that intersect at a 90° angle. Their slopes are negative
reciprocals.
Prior Knowledge
Slope-intercept form: y = mx + b
Standard Form: Ax + By = C
A, B and C are integers (not fractions)
A is positive.
On the left, write the x term first and then the y term. The constant term is written on the right.
New Concepts
1.
Use point-slope form to write the equation of a line.
y – y1 = m(x – x1)
This is the form that should be used to write the equation of a line in most cases. x 1 and y1 are the x- and
y-coordinates of a known point; m is the slope; x and y are the x and y in the final equation.
Example 1: Write the equation of the line through (-3, 5) with a slope of
2
.
5
y - y1 = m(x - x1)
x1 = -3
y1 = 5
2
m=
5
y - __ = __ (x - __ )
2
y  5  x  3
5
This is an equation for the line, but it is not in customary form. Manipulate the equation so that it is in
either slope-intercept form or standard form.
Slope-Intercept Form
Clear the fraction by multiplying each term by 5.
2
y  5   x  3
5
2
5y   55  5 x  3
 5
V. Zabrocki 2011
Standard Form
Clear the fraction by multiplying each term by 5.
2
 x  3
5
2
5y   55  5 x  3
 5
y5 
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M098
Carson Elementary and Intermediate Algebra 3e
Section 4.5
5y  25  2x  3
5y  25  2x  3
5y  25  2 x  6
5y  25  2 x  6
5y  2 x  31
5y  2 x  31
y
 2 x  5y  31
2
31
x
5
5
2 x  5y  31
Example 2: Write the equation of the line through (3, -5) with a slope of 2 in both slope-intercept form
and standard form.
Slope-Intercept Form
y - y1 = m(x - x1)
y - -5 = 2(x – 3)
y + 5 = 2x – 6
y = 2x - 11
Standard Form
y - y1 = m(x - x1)
y - -5 = 2(x – 3)
y + 5 = 2x – 6
-2x + y = -11
2x – y = 11
Example 3: Write the equation of the line through (-1, -5) and (-4, 1) in both slope-intercept form and
standard form.
To use point-slope form, the slope and a point on the line must be known. In this case, there
are two known points but no slope. Use the slope formula to find the slope and then proceed
as before. Either point may be used in the point-slope form.
y  y1
1  5
1 5
6
m  2



 2
x 2  x1
 4  1  4  1  3
Slope-Intercept Form
y - y1 = m(x - x1)
y - -5 = -2(x - -1)
y + 5 = -2(x + 1)
y + 5 = -2x – 2
y = -2x - 7
Standard Form
y - y1 = m(x - x1)
y - -5 = -2(x - -1)
y + 5 = -2(x + 1)
y + 5 = -2x – 2
2x + y = -7
Example 4: Write the equation of the line through (2, -4) and (3, -4) in both slope-intercept form and
standard form.
m 
y 2  y1
4  4
4  4
0



 0
x 2  x1
32
1
1
Because the slope of the line is 0, the line is horizontal. A horizontal line intersects only the yaxis so the equation will have only a y in it. y equals the y-coordinate.
y = -4
Example 5: Write the equation of the line through (-2, 5) and (-2, -6) in both slope-intercept form and
standard form.
m 
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y 2  y1
6  5
11
11



 undefined
x 2  x1
 2  2
22
0
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M098
Carson Elementary and Intermediate Algebra 3e
Section 4.5
Because the slope of the line is undefined, the line is vertical. A vertical line intersects only the
x-axis so the equation will have only an x in it. x equals the x-coordinate.
x = -2
2. Write the equation of a line parallel or perpendicular to a given line.
Parallel lines have equal slopes.
The slopes of perpendicular lines are negative reciprocals.
Example 6: Determine if the lines are parallel or perpendicular.
y = 3x + 2
y = 3x - 2
These lines are written in slope-intercept form (y=mx + b) so the
coefficient of the x-term is the slope. They are both 3 so the lines are
parallel.
Example 7: Determine if the lines are parallel or perpendicular.
5x – 3y = 11
3x + 5y = 8
These lines are written in standard form. Rewrite them in slope
intercept form to find the slope.
5x – 3y = 11
-3y = -5x + 11
5
11
5
y  x
The slope is .
3
3
3
3x + 5y = 8
5y = -3x + 8
3
8
y   x
5
5
The slope is 
3
.
5
5
3
and 
are negative reciprocals so the lines are perpendicular.
3
5
Remember the shortcut. When an equation is written in standard Ax + By = C form,
m = – A/B.
Example 8: Write the equation of the line through (-2, 1) and parallel to y = -3x + 1 in both slopeintercept form and standard form.
To use point-slope form, the slope and a point on the line must be known. In this case,
there is a known point but no slope. Because the desired line is parallel to the given
line, the slope will be the same. So find the slope of the given line.
y = -3x + 1 is written in slope-intercept form (y = mx + b) so the coefficient of the x-term
is the slope. The slope is -3.
Now use the point-slope form as we did earlier to write the equation of the line.
V. Zabrocki 2011
Slope-Intercept Form
Standard Form
y = -3x – 5
3x - y = -5
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M098
Carson Elementary and Intermediate Algebra 3e
Section 4.5
Example 9: Write the equation of the line through (3, 7) and perpendicular to 5x + 2y = 3 in both slopeintercept form and standard form.
To use point-slope form, the slope and a point on the line must be known. In this case,
there is a known point but no slope. Because the desired line is perpendicular to the
given line, the slope will be the negative reciprocal. Find the slope of the given line.
5x + 2y = 3 is written in standard form so rewrite it in slope-intercept form (y = mx + b)
to determine the slope. (Or use the shortcut to determine the slope!)
5x + 2y = 3
2y = -5x + 3
5
3
y   x
2
2
The slope of the given line is 
reciprocal or
5
so the slope we need to use is the negative
2
2
.
5
Now use the point-slope form as we did earlier to write the equation of the line.
Slope-Intercept Form
y 
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2
29
x
5
5
Standard Form
2x – 5y = -29
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