Download Section 4.5 - MSU Billings

4.5
Point-Slope Form
1. Use point-slope form to write the equation of a line.
2. Write the equation of a line parallel to a given line.
3. Write the equation of a line perpendicular to a given line.
Practice ……
1. Write in standard form:
2. Write in standard form:
y  3x  5
y  4x  5
3x  y  5
 4 x  y  5
3. Write in standard form:
2
y  x4
3
3y  2 x  12
 2 x  3y  12
2 x  3y  12
4x  y  5
4. Write in standard form:
5y  2x  6
 2x  5y  6
2x  5y  6
Practice ……
1. Write in standard form:
2
x  5y  6
3
2x  15y  18
3. Write in slope-intercept form:
2
x  5y  6
3
2. Write in standard form:
5  2x  3
2  2x
x 1
4. Write in slope-intercept form:
6x  5 y  2
2 x  15y  18
15y  2 x  18
2
18
x
15
15
2
6
y
x
15
5
y
 5y  6 x  2
y
6
2
x
5
5
Objective 1
Use point-slope form to write the
equation of a line.
Point Slope Form
Given two points (x1, y1) and (x, y)
m
y  y1
x  x1
y  y1  mx  x1 
Use whenever you want to write an equation!
Need to know a point and a slope.
Write the equation of a line with a slope of 5 and passing
through the point (3, 12) in slope-intercept form.
Point: (3, 12)
y – y1 = m(x – x1)
m = 5, x1 = 3, y1 = 12
y – y1 = m(x – x1)
y – 12 = 5(x – 3)
y – 12 = 5x – 15
y = 5x – 3
Slope: 5
Distribute.
Add 12 to both sides to isolate y.
Write the equation in standard form.
y – 12 = 5x – 15
-5x + y -12 = -15
-5x + y = -3
5x – y = 3
Write the equation of a line passing through (-1, -5) and
(-4, 1) in slope-intercept and standard form.
y – y1 = m(x – x1) Point: (-1, -5) Slope:
slope-intercept
y  5  2x  1
y  5  2x  1
y  5  2 x  2
y  5  2x  2
y  2x  7
1  5
6

 2
 4  1  3
standard form
y  5  2x  2
2x  y  7
2
Write the equation of a line with a slope of 5 and
passing through the point (-3, 5) in slope-intercept and
standard form.
y – y1 = m(x – x1)
y5
2
x  3
5
2
 x  3
5
2
5y   55  5 x  3
 5
y5
5y  25  2x  3
5y  25  2x  6
Point: (-3, 5)
slope-intercept
2
Slope:
5
standard form
5y  25  2 x  6
5y  25  2 x  6
5y  2 x  31
 2 x  5y  31
2
31
y x
5
5
2 x  5y  31
Write the equation of a line passing through (2, -4) and
(3, -4) in slope-intercept and standard form.
y – y1 = m(x – x1)
Point: (2, -4) Slope:
Horizontal line
Intersects only the y-axis.
Equation has only a y.
y = -4
m
 4  4
0
32
Write the equation of a line passing through (-2, 5) and
(-2, -6) in slope-intercept and standard form.
y – y1 = m(x – x1) Point: (-2, 5) Slope:
Vertical line
Intersects only the x-axis.
Equation has only an x.
x = -2
m
65
 11

 2  2
0
Objective 2 & 3
Write the equation of a line parallel
to a given line.
Write the equation of a line
perpendicular to a given line.
Parallel Lines
The slopes of parallel lines
are equal.
y = 2x + 1 y = 2x – 3
Perpendicular Lines
The slopes of perpendicular lines are
negative reciprocals.
3
y   x 1
2
a
m
b
m  
b
a
y
2
x4
3
The only way to
determine if lines
are parallel or
perpendicular is to
compare the slopes.
Parallel or Perpendicular?
y  3x  2
m=3
m=3
y  3x  2
Parallel
5 x  3 y  11
3x  5 y  8
y  5x  2
y  5 x  7
m=5
m = -5
Neither
5
3
3
m  
5
m 
Perpendicular
y6
x  5
horizontal
vertical
Perpendicular
Write the equation of a line that passes through (1, –5)
and parallel to y = –3x + 4. Write the equation in
slope-intercept form.
y – y1 = m(x – x1)
Need a point and slope
(1, -5)
y – y1 = m(x – x1)
y – (5) = –3(x – 1)
y + 5 = –3x + 3
y = –3x – 2
m = -3
Write the equation of a line that passes through (3, 7)
and perpendicular to 5x + 2y = 3. Write the equation
in standard form.
y – y1 = m(x – x1)
Need a point and slope
(3, 7)
2
 x  3
5
5y  35  2x  3
y 7 
5y  35  2 x  6
 2 x  5y  29
2 x  5y  29
m 
2
5
Write the equation of the line in
slope-intercept form given m = 2 and the
point (4, 5).
a) y = 2x – 3
b) y = 2x + 3
c) y = 2x + 3
d) y = 2x – 5
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 18
Write the equation of the line in slopeintercept form given m = 2 and the point
(4, 5).
a) y = 2x – 3
b) y = 2x + 3
c) y = 2x + 3
d) y = 2x – 5
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 19
What is the equation of the line connecting
the points (4, 3) and (1, 7) in slopeintercept form?
a) y = 2x + 5
b) y = 2x + 5
c) y = 2x – 5
d)
1
y  x5
2
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 20
What is the equation of the line connecting
the points (4, 3) and (1, 7)?
a) y = 2x + 5
b) y = 2x + 5
c) y = 2x – 5
d)
1
y  x5
2
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 21
What is the relationship between the two
lines?
5x – 3y = 11
3x + 5y = 8
a) parallel
b) perpendicular
c) neither
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 22
What is the relationship between the two
lines?
5x – 3y = 11
3x + 5y = 8
a) parallel
b) perpendicular
c) neither
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 4- 23