Transcript Slopes and Lines

Date:
Topic: Lines and Slope
(1.2)
Definition of Slope
The slope of the line through the distinct points (x1, y1) and (x2, y2) is
Change in y Rise
y2 – y1
=
=
Change in x Run
x2 – x1
where x2 – x1 = 0.
y
Run
x2 – x1
y2
Rise
y2 – y1 y1
(x2, y2)
(x1, y1)
x1
x2
x
Find the slope of the line with points (3, 2) and (-5, 5)
3
5   2 


slope 
5   3 8
The Possibilities for a Line’s Slope
Zero Slope
Positive Slope
y
y
m>0
m=0
x
x
Negative Slope
y
Undefined Slope
y
Line rises from left to right.
Line is horizontal.
m<0
x
Line falls from left to right.
m is
undefined
m= 2
0
Line is vertical.
x
Slope-Intercept Form of the Equation
of a nonvertical line with slope m and y-intercept b is
y = mx + b
Find the slope and the y-intercept of the line whose equation is
2x – 3y + 6 = 0.
2x – 3y + 6 = 0
+ 3y
+3y To isolate the y-term, add 3y on both sides.
2x + 6 = 3y
3y = 2x + 6 Reverse the two sides. (This step is optional.)
3
3
Divide both sides by 3.
y = 2/3x + 2
The slope
is 2/3.
The y-intercept
is 2.
Graphing y=mx+b Using the Slope and Y-Intercept
• Plot the y-intercept, b.
• Plot a second point using the slope, m, rise over run.
• Draw a line through the two points.
Graph the line whose equation is:
y = 2/3 x + 2
5
4
3
2
1
The slope
is 2/3.
The yintercept is 2.
-5 -4 -3 -2
-1
-1
-2
1
2
3 4
5
-3
-4
-5
We need two points in order to graph the line:
We can use the y-intercept, 2, to obtain the first point (0, 2).
We plot the second point on the line by starting at (0, 2), the first
point. Then move 2 units up (the rise) and 3 units to the right (the
run). This gives us a second point at (3, 4).
Point-Slope Form of the equation of a nonvertical line
of slope m that passes through the point (x1, y1) is
two points find the slope using the points,
y – y1 = m(x – x1) Ifandgiven
use one of the coordinates in the equation
Write the point-slope form of the equation of the line passing
through (-1,3) with a slope of 4. Then solve the equation for y.
Solution We use the point-slope equation of a line with m = 4, x1= -1, and
y1 = 3.
y – y1 = m(x – x1)
This is the point-slope form of the equation.
y – 3 = 4[x – (-1)]
Substitute the given values.
y – 3 = 4(x + 1)
We now have the point-slope form of the equation for the given line.
y – 3 = 4x + 4
We can solve the equation for y by applying the distributive property.
+3
+3
y = 4x + 7
Add 3 to both sides.
Equations of Lines
Point-slope form:
Slope-intercept form:
Horizontal line:
Vertical line:
General form:
y – y1 = m(x – x1)
y = mx + b
y=b
x=a
Ax + By + C = 0
Complete Student Checkpoint
Indicate whether each line
has a positive, negative,
zero, or undefined slope.
Write the equation of
the line in slope-intercept
form. (Assume scale is 1)
y-axis
y-axis
undefined
a
zero
x-axis
b
d
-3
slope
-1/2
6
y-intercept
3
x-axis
c
equation:
1
y x3
2
Use the given conditions to write an equation in
point-slope form and slope-intercept form.
Passing through (-3,2) and (3,6). Send you answer
to me using the calculator
y  y1  m(x  x1 )
62
y  (2) 
(x  (3))
3  (3)
2
y  2  (x  3)
3
2
y2 x2
3
2
y x4
3
DAY 2
Slope and Parallel Lines
If two nonvertical lines are parallel, then they have the same slope.
Write an equation of the line passing through (-3, 2) and parallel to
the line whose equation is y = 2x + 1. Express the equation in pointslope form and y-intercept form.
y = 2x + 1
5
(-3, 2)
y – y1 = m(x – x1)
m =2
1
x1 = -3
Parallel lines have the same slope. Because the slope
of the given line is 2, m = 2 for the new equation.
y – 2 = 2[x – (-3)]
y – 2 = 2(x + 3)
y – 2 = 2x + 6
y = 2x + 8
3
Run = 1
2
-5 -4 -3 -2
y1 = 2
4
-1
-1
-2
Rise = 2
1
2
3 4
-3
-4
-5
Apply the distributive property.
This is the slope-intercept form of the equation.
5
Slope and Perpendicular Lines
Two lines that intersect at a right angle (90°) are
said to be perpendicular. There is a relationship
between the slopes of perpendicular lines.
90°
Slope and Perpendicular Lines
• If two nonvertical lines are perpendicular, then the
product of their slopes is –1.
(2/3) • (-3/2) = -1
•
Slopes are negative reciprocals of each other
Find the slope of any line that is perpendicular to the line whose equation is
2x + 4y – 4 = 0.
4y = -2x + 4
y = -1/2x + 1
Slope is –1/2.
Any line perpendicular to this line has a slope that is the negative reciprocal, 2.
You Try
Write an equation of the line
passing through (-2,5) and
parallel to the line whose
equation is y=3x+1. Express
in point-slope form and
slope-intercept form.
y  y1  m(x  x1 )
y  5  3(x  (2))
y  5  3x  6
y  3x  11
Write an equation of the line
passing through (-3,6) and
perpendicular to the line whose
equation is y=1/3 x +4
Express in point-slope form and
slope-intercept form.
3
perpendicular slope: 
 3
1
y  y1  m(x  x1 )
y  6  3(x  (3))
y  6  3x  9
y  3x  3
Graphs and Viewing Windows
•
Bob purchased a house 8 years ago for $42,000.
This year is was appraised at $67,500.
a) A linear equation V=mt + b, 0 ≤ t ≤ 15,
represents the value V of the house for 15
years after it was purchased. Determine m
and b.
b) Graph the equation and trace to estimate in
how many years after the purchase of this
house it will be worth $72,500
c) Write and solve an equation algebraically to
determine how many years after purchase this
house will be worth $74,000.