Transcript Ch 5 Alg1 07-08 UH,MY

CHAPTER V
Writing Linear Equations
By:
Uri Hong
Michael Yanoska
Table of Contents
• 5-1 Slope-Intercept Form
• 5-2 Point-Slope Form
• 5-3 Writing Linear Equations given 2
points
• 5-4 Standard Form
• 5-5 Modeling with Linear Equations
• 5-6 Perpendicular lines
Introduction
This chapter is about writing linear
equations in a variety of algebraic forms
including: slope-intercept form, pointslope form, and standard form.
In this chapter, you will also use a linear
model to solve equations and you will
learn to write an equation perpendicular
to another line.
5-1 Slope-Intercept Form
y=mx+b (slope intercept form)
m= the slope
b=y-intercept
Finding the Slope (m)
When given a graph, find two points on
the graph (x1,y1)(x2,y2).
To find the slope (m):
Rise/Run= (y2-y1)/(x2-x1)
To Find b (y-intercept)
Look at the graph, and see where the
graph crosses the y-axis.
Example (slope-intercept)
(x1,y1)(x2,y2)
(0 , 0)(3, 2 )
(y2 -y1)/(x2 -x1)= m
m=(2-0)/(3-0)
m = 2/3
Y-intercept at (0,0)
Plug into y =mx+b
y=(2/3)x + 0
5-2 Point-Slope Form
y-y1=m(x-x1)
You can use point-slope form to find a
linear equation in slope-intercept form
using the slope m and coordinates that
are on the line.
Example
Write in slope-intercept form the equation of the
line that passes through the point (-3,7) with
slope -2
1. y-y1=m(x-x1)
2. y-7= -2[x-(-3)]
3. y-7=2x-6
4. y = -2x+1
Just a side note (:
When given a question asking to find an
equation parallel, it means that the 2
equations will have the same slope
y= 2x+3 and y= 2x+9 are parallel
When given a question asking to find an
equation perpendicular, it means that the 2
equations will have slopes that are opposite
reciprocals.
y= -2x +3 and y= (-1/2)x=6 are perpendicular.
Writing Linear Equations given
2 Points
Write in slope-intercept form the equation
of the line that passes through the
points (3,-2) and (6,0)
1. Find the slope. Use (x1,y1)=(3,-2)
(x2,y2)=(6,0)
m=(y2-y1)/(x2-x1)
[0-(-2)]/(6-3) =m= 2/3
(x1,y1)=(3,-2)
y-y1=m(x-x1) (point-slope form)
y-(-2)=(2/3)(x-3)
y+2=(2/3)x-2
Y=(2/3)x-4
Standard Form
The standard form of an equation of a line
is
Ax+By=C
A and B = coefficients ≠0
Example
Write y=(2/5)x-3 in standard form with integer
coefficients.
Ax+By=C
y=(2/5)x-3
multiply each side by 5
5y=5[(2/5)x-3]
Distribute
5y=2x-15
Subtract 2 from each side -2x+5y=-15
Rewrite with leading Coefficient positive
Hence: 2x – 5y = 15
Modeling with Linear
Equations
A linear model = simulate a real life
situation.
The rate of change =compares the two
entities that are changing.
The slope = rate of change
From 1990- 2008 the number of McDonalds in
the U.S. increase by about 500 per year. In
2000, there were about 20,000 McDonalds.
Write a linear model expressing the number
of McDonalds.
Let t=0 represent 1990
m=500 (slope= rate of increase)
t=10 (the year 2000)
Therefore, (t1,r1)= (10,20000)
y-r1=m(t-t1)
y-20000=500(t-10)
y-20000=500t-5000
y=500t+15000
Perpendicular Lines
Two lines are perpendicular if the 2 lines
intersect and form a 90° angle.
When given a question asking to find an
equation perpendicular, it means that the 2
equations will have slopes that are
reciprocals.
y=2x +3 and y=(-1/2)x+6
are perpendicular.
Find the slope that is perpendicular to the
equation that has the following points:
(3,2)(6,4) .
1. Find the slope: (y2-y1)/(x2-x1)
(4-2)/(6-3)= 2/3
2. Point-slope form : y-y1=m(x-x1)
y-2=(2/3)(x-3)
y-2=2/3x-2
y=2/3x
3. Find the slope perpendicular:
m=2/3
reciprocal= 3/2
take the opposite= -(3/2)
slope of the line perpendicular = -(3/2)
Summary
•
•
•
•
•
5-1 Slope-Intercept Form (y=mx+b)
5-2 Point-Slope Form (y-y1=m(x-x1))
5-3 Writing Linear Equations given 2 points
5-4 Standard Form (Ax+By=C)
5-5 Modeling with Linear Equations (real-life
models)
• 5-6 Perpendicular lines (opposite-reciprocal
slopes)