1. After today, the next four class periods are:
Review for Quiz 2
Quiz 2 (on sections )
Review for Test 1
Test 1 (Chapters 1, 2, 3)

2. Any questions on the Section 3.4 homework?

3. CLOSE Please YOUR LAPTOPS, and get out your note-taking materials.
and turn off and put away your cell phones,
and get out your note-taking materials.

4. Section 3.5 Equations of Lines
Recall that the slope-intercept form of a line is
y = mx + b ,
where the line has a slope of m
and has a y-intercept of (0, b).
If we know the slope and y-intercept of a line, we can substitute into this form to get an equation for the line.

5. Example
Find an equation of a line with slope of -3 and y-intercept of (0, ).
By substituting the appropriate values into the slope-intercept form, we get
y = -3x – .
Note: If you’re asked to write the equation in standard form, the answer could be converted to 3x + y =
If you prefer to not use fractions in your final answer, multiply by 5 to get 15x + 5y = -1. (This is the way “standard form” is usually written, w/o fractions.)

6. Example
Graph y = x – 2.
We can use the slope-intercept form to help us graph the equation.
We know that the y-intercept is (0, -2), which gives us one point for the line.
We can also use the definition of slope to help us get another point.
slope = , which in this case is

7. Example (cont.) First we graph the y-intercept. y
Then we use the slope of -4/5 to find another point.
Move down 4 and to the right 5.
(0, -2)
4 units down
This gives us the new point (5, -6).
(5, -6)
5 units right
Now draw the line.

8. The slope-intercept form uses, specifically, the y-intercept in the equation. The point-slope form allows you to use ANY point, together with the slope, to form the equation of the line. Point-slope formula for linear equations:
Where m is the slope, and
(x1, y1) is a point on the line

9. ATTENTION! Pay special attention to this next slide and the examples that follow, because SEVERAL problems on both Quiz 2 and Test 1 will use this formula!

10. y + 12 = -2x – 22 (use distributive property)
Example
Find an equation of a line with slope –2, through the point (-11,-12). Write the final equation in slope-intercept form.
(Note: it’s always a good idea to graph the line first. This will help you see if your equation makes sense, which is especially helpful on quizzes and tests.
Solution: Substitute the slope and point into the point-slope form of the linear equation:
y – (-12) = -2(x – (-11))
y + 12 = -2x – (use distributive property)
y = -2x (subtract 12 from both sides)
So the slope is -2, and the y-intercept is (0,-34)

11. Problem from today’s homework:

12. Problem from today’s homework:

13. Example
Find the equation of the line through (-4,0) and (6,-1). Write the equation in standard form.
First find the slope.

14. Example (cont.)
Now substitute the slope and one of the points into the point-slope form of an equation.
(clear fractions by multiplying both sides by 10)
(use distributive property)
(add x to both sides)
NOTE: In slope-intercept form, this would be y = - 1 x - 2

15. Problem from today’s homework:

16. Example
Find the equation of the line passing through points (2, 5) and (-4, 3). Write the equation using function notation.
First, calculate the slope:

17. Example (cont.) Now enter the slope and one of the points
(either one will work) into the point-slope
equation:
(Looks just like slope-intercept form,
but with f(x) instead of y.)

18. Problem from today’s homework:

19. Example
Find the equation of the horizontal line through (1, 4).
Recall that horizontal lines have an equation of the form y = c.
So using the y-coordinate in the given point, y = 4.

20. Example
Find the equation of the vertical line through (-1,3).
Recall that vertical lines have an equation of the form x = c.
So using the x-coordinate in the given point, x = -1.

21. Problem from today’s homework:

22. Word problem similar to final problems in today’s homework:

23. 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, you can easily determine slope to compare lines.

24. Example:
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.
First, we need to find the slope of the given line.
3y = -x (subtract x from both sides)
y = x (divide both sides by 3)
Since parallel lines have the same slope, we use the slope of for our new equation, together with the given point.

25. Example (cont.) (multiply by 3 to clear fractions)
(use distributive property)
(add x to both sides) (Why? Because they want it in STANDARD form)
(add 12 to both sides)
What would this look like in slope-intercept form?
In function notation?

26. 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 (subtract 3x from both sides)
(divide both sides by 2)
Since perpendicular lines have slopes that are negative reciprocals of each other, we use the slope of for our new equation, together with the given point (3,-5).

27. Example (cont.) (multiply by 3 to clear fractions)
(use distributive property)
(subtract 15 from both sides)
(divide both sides by 3)
What would this look like in function notation?

28. Problem from today’s homework:

29. The line y = -4 is a horizontal line (slope = 0).
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.

30. 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.

31. Problem from today’s homework:

32. Problem from today’s homework:

33. The homework assignment
Reminder:
The homework assignment
on Section 3.5 is due
at the start of
next class period.

34. and begin working on the homework assignment.
You may now
OPEN
your LAPTOPS
and begin working on the homework assignment.