Transcript Section 1.5 - El Camino College

Section 1.5
Finding Linear Equations
Using Slope and a Point to Find an Equation of a Line
Method 1:
Using Slope Intercept
Example
Find an equation of a line that has slope m = 3 and
contains the point (2, 5).
Solution
• Substitute m  3 into the slope-intercept form:
y  3 x  b.
• Now we must find b
• Every point on the graph of an equation represents
of that equation, we can substitute x  2 and y  5
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 2
Using Slope and a Point to Find an Equation of a Line
Method 1:
Using Slope Intercept
Solution Continued
5 = 3(2) + b
Substitute 2 for x and 5 for y.
5=6+b
Multiply.
5–6=6+b–6
–1=b
Subtract 6 from both sides.
Simplify.
We now substitute –1 for b into y = 3x + b:
y = 3x – 1
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 3
Using Slope and a Point to Find an Equation of a Line
Method 1:
Using Slope Intercept
Graphing Calculator
We can use the TRACE on a graphing calculator
to verify that the graph of y  3x  1 contains the
point (2, 5).
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 4
Using Two Points to Find an Equation of a Line
Method 1:
Using Slope Intercept
Example
Find an equation of a line that contains the points (–2,
6) and (3, –4).
Solution
• Find the slope of the line:
4  6
10 10
m


 2
3  (2) 3  2
5
• We have y  2 x  b
• Line contains the point (3, –4)
• Substitute 3 for x and –4 for y
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 5
Using Two Points to Find an Equation of a Line
Method 1:
Using Slope Intercept
Example
–4 = –2(3) + b Substitute 3 for x. –4 for y.
–4 = –6 + b
Multiply.
–4 + 6 = –6 + b + 6 Add 6 to both sides.
2=b
Simplify.
•Substitute 2 for b into y = –2x + b:
y = –2x + 2
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 6
Using Two Points to Find an Equation of a Line
Method 1:
Using Slope Intercept
Graphing Calculator
We can use the TRACE on a graphing calculator to
verify that the graph of y = –2x + 2 contains the
points (–2, 6) and (3, –4).
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 7
Finding a Linear Equation That Contains Two Given Points
Method 1:
Using Slope Intercept
Guidelines
To find the equation of a line that passes through
two given points whose x-coordinates are
y2  y1
different,
m
x2  x1
1. Use the slope formula,
, to find the
slope of the line.
2. Substitute the m value you found in step 1 into
the equation y  mx  b.
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 8
Finding a Linear Equation That Contains Two Given Points
Method 1:
Using Slope Intercept
Guidelines Continued
3. Substitute the coordinates of one of the given
points into the equation you found in step 2, and
solve for b.
4. Substitute the m value your found in step 1 and the
b value you found in step 3 into the
equation y  mx.  b.
5. Use a graphing calculator to check that the graph
of your equation contains the two points.
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 9
Using Two Points to Find an Equation of a Line
Method 1:
Using Slope Intercept
Example
Find an equation of a line that contains the points (–3,
–5) and (2, –1).
Solution
First we find the slope of the line:
1   5  1  5 4
m


2  (3)
23 5
4
5
• We have y = x + b.
• The line contains the point (2, –1)
• Substitute 2 for x and –1 for y:
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 10
Using Two Points to Find an Equation of a Line
Method 1:
Using Slope Intercept
Solution Continued
4
–1 = (2) + b Substitute 2 for x. –1 for y.
5
4 2
4
4
8
–1 = + b
(2) =5 ∙1 =
5
5
5
8
5∙(–1) = 5∙ 5 + 5∙b Multiply both sides by 5.
8 5 8 8
–5 = 8 + 5b
5∙ = ∙ = = 1
5 1 5 1
–13 = 5b
Subtract 8 from both sides.
13
– =b
Divide both sides by 5.
.5
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 11
Using Two Points to Find an Equation of a Line
Method 1:
Using Slope Intercept
Solution Continued
13
4
So, the equation is y = x – .5
5
Graphing Calculator
We can use the TRACE on a
graphing calculator to verify that
4 13
the graph of y  x  contains
5
5
the points (–3, –5) and (2, –1).
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 12
Finding an approximate Equation of a Line
Method 1:
Using Slope Intercept
Example
Find an approximate equation of a line that contains
the points (–6.81, 7.17) and (–2.47, 4.65). Round the
slope and the constant term to two decimal places.
Solution
First we find the slope of the line:
4.65  7.17
2.52
m

 0.58
2.47   6.81 4.34
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 13
Finding an approximate Equation of a Line
Method 1:
Using Slope Intercept
Solution Continued
We have y = –0.58x + b. Since the line contains the
point (–6.81, 7.17), we substitute –6.81 for x and
7.17 for y:
Sub -6.81 for x, 7.17
7.17 = –0.58(–6.81) + b
7.17 = 3.9498 + b
for y.
Multiply.
7.17 – 3.8498 = 3.9498 + b – 3.9498 Subtract 3.9498 from
3.22  b
Section 1.5
both sides.
Combine like terms.
Lehmann, Intermediate Algebra, 3ed
Slide 14
Finding an approximate Equation of a Line
Method 1:
Using Slope Intercept
Solution Continued
So, the equation is y = –0.58x + 3.9498
Graphing Calculator
We can use the TRACE on a
graphing calculator to verify that the
graph of y  0.58 x  3.9498 comes
very close to the points (–6.81, 7.17)
and (–2.47, 4.65).
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 15
Finding an Equation of a Line Parallel to a Given Line
Method 1:
Using Slope Intercept
Example
Find an equation of a line l that contains the point
(5, 3) and is parallel to the line y = 2x – 3.
Solution
• Line y = 2x – 3 the slope is 2
• The parallel line l has slope 2
• So, y = 2x + b
• To find b we substitute 5 for x and 3 for y:
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 16
Finding an Equation of a Line Parallel to a Given Line
Method 1:
Using Slope Intercept
Solution Continued
3 = 2(5) + b Substitute 5 for x and 3 for y.
–7 = b
Multiply, subtract by 10 from both sides.
So, the equation of l is y = 2x – 7.
Graphing Calculator
We can use the TRACE on a
graphing calculator to verify our
equation.
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 17
Finding an Equation of a Line Perpendicular to a Given Line
Method 1:
Using Slope Intercept
Example
Find an equation of a line l that contains the point
(2, 5) and is perpendicular to the line –2x + 5y = 10
Solution
First we isolate y:
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 18
Finding an Equation of a Line Perpendicular to a Given Line
Method 1:
Using Slope Intercept
Solution Continued
2
2
For the line y = x + 2, the slope is m = . The
5
5
slope of the line l must be the opposite of the
2
5
5
reciprocal , or – . An equation of l is y = – x + b.
5
2
2
To find b, substitute 2 for x and 5 for y.
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 19
Finding an Equation of a Line Perpendicular to a Given Line
Method 1:
Using Slope Intercept
Solution Continued
5
The equation of l is y = – x + 10.
2
Graphing Calculator
Use ZStandard followed by
ZSquare to verify our work.
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 20
Finding an Equation of a Line Perpendicular to a Given Line
Method 1:
Using Slope Intercept
Example
Find an equation of a line l that contains the point (4,
3) and is perpendicular to the line x = 2.
Solution
• Graph of x = 2 is a vertical line
• A line perpendicular to it must be
horizontal
• There is an equation of l of the
form y = b.
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 21
Finding an Equation of a Line Perpendicular to a Given Line
Method 1:
Using Slope Intercept
Solution Continued
• We substitute the ycoordinate of the given
point (4, 3) into y = b
•3=b
• An equation of l is y = 3.
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 22
Defining Point-Slope Form
Method 2:
Using Point-Slope
Second method to find a linear equation of a line.
Suppose that a nonvertical line has:
• Slope is m
• y-intercept is (x1, y1)
• (x, y) represents a different point on the line
y  y1
So, the slope is:
m
x  x1
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 23
Defining Point-Slope Form
Method 2:
Using Point-Slope
Given the slope, multiple both sides by x – x1 gives
y  y1
∙(x – x1) = m (x – x1)
x  x1
y – y1 = m (x – x1)
We say that this linear equation is in point-slope form.
IfDefinition
a nonvertical line has slope m and contains the point
(x1, y1), then an equation of the line is
y – y1 = m (x – x1)
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 24
Using Point-Slope Form to Find an Equation of a Line
Method 2:
Using Point-Slope
Example
A line has slope m = 2 and contains the point (3, –8).
Find the equation of the line
Solution
Substituting x1 = 3, y1 = –8 and m = 2 into the
equation y – y1 = m (x – x1).
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 25
Using Point-Slope Form to Find an Equation of a Line
Method 2:
Using Point-Slope
Example
Use point-slope form to find an equation of the line
that contains the points (–5, 2) and (3, –1). Then
write in slope-intercept form.
Solution
First find the slope of the line:
1  2 3
3
m


3   5  8
8
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 26
Using Point-Slope Form to Find an Equation of a Line
Method 2:
Using Point-Slope
Solution Continued
3
Substituting x1 = 3, y1 = –1 and m =  into the
8
equation y – y1 = m (x – x1).
Section 1.5
Lehmann, Intermediate Algebra, 3ed
Slide 27