Transcript 3-5 Finding an Equation of a Line

Standard Form
Ax + By = C
3x – 4y = 24
x + 2y = 6
2x – 5y = –10
x=–7
Slope
3/4
y-intercept
–6
–1/2
3
2/5
2
undefined
none
x-intercept
8
6
–5
–7
y – y1 = m (x – x1)
m = slope
x1 = x-coordinate of point on line
y1 = y-coordinate of point on line
To convert to Slope-intercept Form:


Distribute the slope.
Move y1 to right side of the equal sign.
Example
4/8/2016
Give an equation in Standard Form
3
for the line that has a slope of 2
passes through the point (8, –2).
Find the x-int and y-int.
Slope = – 2
and (8, –2)
y  y1  m( x  x1 )
3
Substitute your given
y  (2)  ( )( x  (8))
values →
2
3
y  2  ( x  8)
2
3
y  2  x  12
Distribute →
2
Point-Slope Form →
Continue →
Continued
4/8/2016
Continued from
3
the previous slide → y  2  x  12
2
3
y

x

14
Solve for y →
2
3
Put into Standard Form →  x  y  14
2
Multiply by – 2 →
Standard Form →
3
 2[ x  y  14]
2
3 x  2 y  28
x-int = 28/3
y-int = –14
Example
4/8/2016
Give an equation in Standard Form
for the line that passes through the
points (3, 5) and (6, – 1)
1 5
slope 

6  3 Point-Slope Form → y  y  m( x  x )
Slope = – 2
Pick one of the
given points.
Let’s pick (3, 5)
1
Substitute your given
values →
1
y  (5)  (2)( x  (3))
y  5  2( x  3)
y  5  2 x  6
y  2 x  11
Put into Standard Form → 2 x  y  11
Distribute →
Solve for y →
Example Continued
4/8/2016
Give an equation in Standard Form
for the line that passes through the
points (3, 5) and (6, – 1)
1 5
slope 

6  3 Point-Slope Form → y  y  m( x  x )
Slope = – 2
Pick one of the
given points.
Let’s pick (6, –1)
1
Substitute your given
values →
1
y  (1)  (2)( x  (6))
y  1  2( x  6)
y  1  2 x  12
y  2 x  11
Put into Standard Form → 2 x  y  11
Distribute →
Solve for y →
Example
4/8/2016
Give an equation in Standard Form
for the line that has an x-int 3 and
y-int 7. The ordered pairs are (3, 0) and (0, 7).
Point-Slope Form → y  y1  m( x  x1 )
70
slope 
 Substitute your given
7
y  (0)  ( )( x  (3))
0  3 values →
3
Slope = – 7/3
Pick one of the
given points.
Let’s pick (3, 0)
Distribute →
Put into Standard Form →
7
y   ( x  3)
3
7
y   x7
7 3
x y 7
3
Continued
→
Continued
Continued from
the previous slide →
7
x y 7
3
Multiply by 3 →
7
3[ x  y  7]
3
Standard Form →
7 x  3 y  21
4/8/2016
Example
4/8/2016
Last week, Mr. Baxter sold $20,000 worth of newspaper
advertisements and earned $800. The week before, he sold
$26,000 worth of advertisements and earned $860. Assume
the relationship between Mr. Baxter’s weekly earnings and
the value of the advertisements he sells is linear.
a) Write the data above as two ordered pairs in the form
(s, e). (20,000, 800) and (26,000, 860)
b) Find the slope of the line given these ordered pairs.
Slope = .01
c) Write an equation for the line in slope-intercept form
given the data.
Continued→
Example
4/8/2016
Last week, Mr. Baxter sold $20,000 worth of newspaper
advertisements and earned $800. The week before, he sold
$26,000 worth of advertisements and earned $860. Assume
the relationship between Mr. Baxter’s weekly earnings and
the value of the advertisements he sells is linear.
Ordered Pairs (20,000, 800) and (26,000, 860)
Slope = .01
c) Write an equation for the line in slope-intercept form given
the data.
y  y1  m( x  x1 )
Slope = .01 and (20,000, 800)
y  800  .01( x  20000)
y  800  .01x  200
y  .01x  600






You cannot find 1 equation for a piecewise
function.
You can find an equation for each part.
Use the endpoints of each part to find the
slope of that segment.
Use the slope and one of the points to write
an equation in point-slope form.
Convert to slope intercept form.
Write an inequality to define that piece.




Use the endpoints of
each part to find
slope.
Write an equation in
point-slope form.
Convert to slopeintercept form.
Write an inequality.
30
3
1 0
y  3  3( x  1)
y  3x
0  x 1




Use the endpoints of
each part to find
slope.
Write an equation in
point-slope form.
Convert to slopeintercept form.
Write an inequality.
43 1

5 1 4
1
y  4  ( x  5)
4
1
11
y  x
4
4
1 x  5




Use the endpoints of
each part to find
slope
write an equation in
point-slope form
Convert to slope
intercept
Write an inequality
04
 4
65
y  0  4( x  6)
y  4 x  24
5 x6
1
11
y  x  ;1  x  5
4
4
y  4 x  24;5  x  6
y  3x;0  x  3
Lesson Master 3-5.
 Pages 165-166 #’s 2, 5, 7-9, 13, 14, 16.
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
Quiz on Sections 3-4 and 3-5 on
Tuesday, November 25th!!!