Please close your laptops
and turn off and put away your
cell phones, and get out your
note-taking materials.
Writing Linear Equations 2
Section 8.1
Review of formulas covered in Chapter 3:
Note: You can use a printed
copy of the sheet containing
formulas on quizzes/tests.
Yellow handout copies for
you to keep are available
here and in the open lab.
(You can also view/print a copy of the
formula sheet online by clicking on
the “Formula sheet” menu button.)
Keep a copy of this sheet in your
notebook to use while you do your
homework assignments and practice
quizzes/tests so you know which
ones will be available to you on the
sheet and where to find them.
Clean copies will be handed out for
your use at each quiz/test, so you may
write notes on your own copy, but
you won’t have those notes available
as you take the quiz in class.
Last session we applied the point-slope formula to problems in
which we were given the slope and one point on the line.
Today we’ll be applying the point-slope formula to problems
with different sets of information:
Example: What if we’re given TWO POINTS on the line?
Find the equation of the line through (-4,0) and
(6,-1). Write the equation in standard form.
• First find the slope.
y2  y1
1 0
1
m


x2  x1 6  (4) 10
y  y1  m( x  x1 )
Example (cont.)
• Now substitute the slope and either one of the
points into the point-slope formula.
1
2
(This would give y  
x
1
10
5
y  0   ( x  (4))
but we need standard form.)
10
10 y  1( x  4)
For standard form, clear fractions
by multiplying both sides by 10.
10 y   x  4
(use distributive property)
x  10 y  4
(add x to both sides)
Problem from today’s homework:
Answer in slope-intercept form:
y = -1/3x +10/3
Answer in standard form:
x + 3y = 10
Quick review:
parallel and perpendicular lines
•
Nonvertical parallel lines have identical
slopes.
•
Nonvertical perpendicular lines have slopes
that are negative reciprocals of each other.
Remember: If you rewrite linear equations
into slope-intercept form (i.e. solve for y),
you can easily determine the slope.
Example
Find an equation of a line that contains the point (3,-5)
and is perpendicular to the line 3x + 2y = 7. Write the
equation in slope-intercept form.
•
First, we need to find the slope of the given line.
2y = -3x + 7
(subtract 3x from both sides)
3
7
y  x
2
2
•
(divide both sides by 2)
Since perpendicular lines have slopes that are negative
reciprocals of each other, we use the slope of 2/3 for our
new equation, together with the given point (3,-5).
Example (cont.)
2
y  (5)  ( x  3)
3
y  y1  m( x  x1 )
(multiply by 3 to clear fractions)
3( y  5)  2( x  3)
(use distributive property)
3 y  15  2 x  6
(subtract 15 from both sides)
3 y  15  15  2 x  6  15
3 y  2 x  21
2
y  x7
3
(simplify)
(divide both sides by 3)
Example:
•
First, we need to find the slope of the given line.
3y = -x + 6
(subtract x from both sides)
y=
•
Find an equation of a line that contains the point
(-2,4) and is parallel to the line x + 3y = 6.
Write the equation in standard form.

1
3
x+2
(divide both sides by 3)
Since parallel lines have the same slope, we use the slope
of  13 for our new equation, together with the given point.
y  y1  m( x  x1 )
Example (cont.)
1
y  4   ( x  (2))
3
3( y  4)  1( x  2)
3 y  12   x  2
(multiply by 3 to clear fractions)
(use distributive property)
x  3 y  12  2 (add x to both sides) (Why? Because they
x  3 y  10
want it in STANDARD form)
(add 12 to both sides)
1
10
What would this look like in slope-intercept form? y   x 
3
3
Example:
Find the equation of the line parallel to y = -4,
passing through the point (0,-3).
• The line y = -4 is a horizontal line (slope = 0).
• If the new line is parallel to this horizontal line
•
•
y = -4, then it must also be a horizontal line.
So we use the y-coordinate of our point to find
that the equation of the line is y = -3.
NOTE: Sketching a quick graph of the line y = -4 and
the point (0,-3) can help you visualize the situation and
make sure you are solving the problem correctly.
Example
Find the equation of the line perpendicular to
x = 7, passing through the point (-5,0).
•
•
•
•
The line x = 7 is a vertical line.
If the new line is perpendicular to the vertical line
x = 7, then it must be a horizontal line.
So we use the y-coordinate of our point to find that the
equation of the line is y = 0.
Again: Sketching a quick graph of the line x = 7 and
the point (-5,0) can help you visualize the situation and
make sure you are solving the problem correctly.
Application problems that involve figuring
out the equation of a line:
The assignment on this material (HW 8.1)
Is due at the start of the next class session.
Lab hours:
Mondays through Thursdays
8:00 a.m. to 6:30 p.m.
You may now OPEN
your LAPTOPS
and begin working on the
homework assignment.