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16
Warm-Up:

3. Write the equation in standard form using integer
coefficients only: 3y = -2/5x + 4
Warm-Up:
5. What is the equation of the line shown?
y

x
Practice Problems:
1. Is the graph a function?
Why or why not?
2. Is the graph a linear function?
Why or why not?
3. What is the per week rate
of change between weeks 1-3?
4. What is the per week rate of
change between weeks 6-10?
Practice Problems:
Parallel & Perpendicular Lines:
Formulas & Vocabulary Section of Notebook, Please
Two lines are considered parallel if they have the
same slope.
Two lines are considered to be perpendicular to each
other if their slopes are the opposite inverse (negative
reciprocal) of each other. Perpendicular lines cross
each other at a 90° angle.
The slope of a perpendicular line is:
Parallel & Perpendicular Lines:
Let's find the equation of a line parallel
to y = - x that passes through the point
(2, 4) What is the slope of the first
line, y = - x ? -1

y = - x (+ 0)

This is in slope intercept form so y =
mx + b which means the slope is –1.

Use the point-slope
formula to find the
equation of the 2nd line
y -y1= -(x - x1)
y -4 = -(x - 2)

y = -x + 6
Practice Problem:
Write an equation of the line that passes through (–3, –5)
and is parallel to the line y = 3x – 1.
SOLUTION
STEP 1
Identify the slope. The graph of the given equation has a
slope of 3. So, the parallel line through (–3, –5) has a
slope of 3.
STEP 2
Find the y-intercept. Use the given point and the slope.
Practice Problem:
y = mx + b
–5 = 3(–3) + b

4=b

Write slope-intercept form.
Substitute 3 for m, 3 for x, and 5 for y.
Solve for b.

STEP 3
Write an equation. Use y = mx + b.
y = 3x + 4

Substitute 3 for m and 4 for b.
Parallel & Perpendicular Lines:
Let's find the equation of a line
perpendicular to y = - x that passes thru
the point (2, 4)
The slope of the first line
is still –1.
The slope of a line perpendicular
is the negative reciporical so
take –1 and "flip" it over and
make it negative.

The slope of a perpendicular
line is 1 and it passes through
(2, 4).

Use the point-slope
formula to find the
equation of the 2nd line.

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January 16, 2014

  • 1. 16
  • 2. Warm-Up: 3. Write the equation in standard form using integer coefficients only: 3y = -2/5x + 4
  • 3. Warm-Up: 5. What is the equation of the line shown? y x
  • 4. Practice Problems: 1. Is the graph a function? Why or why not? 2. Is the graph a linear function? Why or why not? 3. What is the per week rate of change between weeks 1-3? 4. What is the per week rate of change between weeks 6-10?
  • 6. Parallel & Perpendicular Lines: Formulas & Vocabulary Section of Notebook, Please Two lines are considered parallel if they have the same slope. Two lines are considered to be perpendicular to each other if their slopes are the opposite inverse (negative reciprocal) of each other. Perpendicular lines cross each other at a 90° angle. The slope of a perpendicular line is:
  • 7. Parallel & Perpendicular Lines: Let's find the equation of a line parallel to y = - x that passes through the point (2, 4) What is the slope of the first line, y = - x ? -1 y = - x (+ 0) This is in slope intercept form so y = mx + b which means the slope is –1. Use the point-slope formula to find the equation of the 2nd line y -y1= -(x - x1) y -4 = -(x - 2) y = -x + 6
  • 8. Practice Problem: Write an equation of the line that passes through (–3, –5) and is parallel to the line y = 3x – 1. SOLUTION STEP 1 Identify the slope. The graph of the given equation has a slope of 3. So, the parallel line through (–3, –5) has a slope of 3. STEP 2 Find the y-intercept. Use the given point and the slope.
  • 9. Practice Problem: y = mx + b –5 = 3(–3) + b 4=b Write slope-intercept form. Substitute 3 for m, 3 for x, and 5 for y. Solve for b. STEP 3 Write an equation. Use y = mx + b. y = 3x + 4 Substitute 3 for m and 4 for b.
  • 10. Parallel & Perpendicular Lines: Let's find the equation of a line perpendicular to y = - x that passes thru the point (2, 4) The slope of the first line is still –1. The slope of a line perpendicular is the negative reciporical so take –1 and "flip" it over and make it negative. The slope of a perpendicular line is 1 and it passes through (2, 4). Use the point-slope formula to find the equation of the 2nd line.