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**CHAPTER V Writing Linear Equations**

By: Uri Hong Michael Yanoska

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**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

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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.

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**5-1 Slope-Intercept Form**

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

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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.

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**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

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**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.

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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

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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.

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**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

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**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

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

A and B = coefficients ≠0

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**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

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**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

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**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

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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.

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**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)

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**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)

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