Download presentation

1
**CHAPTER V Writing Linear Equations**

By: Uri Hong Michael Yanoska

2
**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

3
Introduction This chapter is about writing linear equations in a variety of algebraic forms including: slope-intercept form, point-slope 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.

4
**5-1 Slope-Intercept Form**

y=mx+b (slope intercept form) m= the slope b=y-intercept

5
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.

6
**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

7
**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.

8
Example Write in slope-intercept form the equation of the line that passes through the point (-3,7) with slope -2 y-y1=m(x-x1) 2. y-7= -2[x-(-3)] 3. y-7=2x-6 4. y = -2x+1

9
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.

10
**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) Find the slope. Use (x1,y1)=(3,-2) (x2,y2)=(6,0) m=(y2-y1)/(x2-x1) [0-(-2)]/(6-3) =m= 2/3

11
**y-y1=m(x-x1) (point-slope form)**

(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

12
**Standard Form The standard form of an equation of a line is Ax+By=C**

A and B = coefficients ≠0

13
**Write y=(2/5)x-3 in standard form with integer coefficients.**

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 y=5[(2/5)x-3] Distribute y=2x-15 Subtract 2 from each side -2x+5y=-15 Rewrite with leading Coefficient positive Hence: 2x – 5y = 15

14
**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

15
**m=500 (slope= rate of increase)**

From 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

16
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.

17
**Find the slope: (y2-y1)/(x2-x1)**

Find the slope that is perpendicular to the equation that has the following points: (3,2)(6,4) . 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)

18
**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)

Similar presentations

Presentation is loading. Please wait....

OK

1-31-13 Unit 1 Basics of Geometry Linear Functions.

1-31-13 Unit 1 Basics of Geometry Linear Functions.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google

Ppt on brand equity management Download ppt on coordinate geometry for class 9th date Pdf to ppt online converter free download Ppt on event handling in javascript all objects Ppt on force and friction for class 8 Ppt on political parties and electoral process in america Ppt on single phase and three phase dual converter travel Ppt on abo blood grouping experiment Ppt on construction of bridges in hilly areas Ppt on social contract theory pdf