Linear Equations ReviewLinear Equations ReviewChapter 5

Chapters 1 & 2

What you should know aboutWhat you should know aboutLinear equationsLinear equations

SlopeY-interceptX-interceptWhat does the graph

look like?Parallel slopePerpendicular slope

Given ANY linear equation you should be able to identify…

Equation FormsEquation Forms

Slope Intercept

StandardHorizontalVertical

y = mx + bAx + By = Cy = bx = a

SlopesSlopesNegativePositive

Horizontal Vertical

Can you run through the Can you run through the linear equation information…linear equation information…

3x+4y=243x+4y=24

y = 1/2x-7y = 1/2x-7

y = 5y = 5

x = 6x = 6

Given any linear equation, one Given any linear equation, one should be able to should be able to 3x+4y=243x+4y=24

The Equation FormDirectionSlopey-interceptx-interceptParallel SlopePerpendicular Slope

1. Standard2. Falling3. -3/4 4. 65. 86. -3/4 7. 4/3

identify…

Given any linear equation, one Given any linear equation, one should be able to should be able to y = 1/2x-7y = 1/2x-7

The Equation FormDirectionSlopey-interceptx-interceptParallel SlopePerpendicular Slope

1. Slope intercept2. Rising3. 1/24. -75. - -7/(1/2) = 146. 1/27. -2

identify…

Given any linear equation, one Given any linear equation, one should be able to should be able to y = 5y = 5

The Equation FormDirectionSlopey-interceptx-interceptParallel SlopePerpendicular Slope

1. Horizontal line2. horizontal3. 04. 55. Does not exist6. 07. undefined

identify…

Given any linear equation, one Given any linear equation, one should be able to should be able to x x = 6 = 6

The Equation FormDirectionSlopey-interceptx-interceptParallel SlopePerpendicular Slope

1. Vertical line2. verticle3. undefined 4. Does not exist5. 66. undefined 7. 0

identify…

To graph a lineTo graph a line

Intercepts◦Identify the intercepts

◦Plot the intercepts

◦Draw the line

Point-slope◦Identify a point on the line and the slope

◦Plot the point◦Count the slope “Rise/Run”

(0,-7)

(-7/3,0)

Graph using intercepts Graph using intercepts y = -y = -3x – 73x – 7

y-int = -7

x-int = 7

-3

up 3 back 1

(0,-7)

down 3 over 1

Graph using intercepts Graph using intercepts y = -y = -3x – 73x – 7

point= (0, -7)

slope = -3 / 1

The slope formulaThe slope formula

m = y1 – y2

x1 – x2

This is really the same at the point-slope equation

m(x1 – x2) = y1 – y2

y1 – y2 = m(x1 – x2)

Find the slope given 2 points (-Find the slope given 2 points (-1,1); (2,3)1,1); (2,3)

m = 3 – 1 2 – -1m = 2 3

m = 1 – 3 -1 – 2m = -2 = 2 -3 3

Now write the equation (-1,1); Now write the equation (-1,1); (2,3)(2,3)

m = 3 – 1 2 – -1

m = 2 3

y1 – y2 = m(x1 – x2)y – 3 = 2/3 (x – 2)y = 2/3 x – 4/3 + 9/3y = 2/3 x + 5/3

If you have two points you can find If you have two points you can find the line…sometimes the challenge is the line…sometimes the challenge is knowing what you have.knowing what you have.

Given◦ The origin◦ The y-intercept◦ The x-intercept◦ A line parallel to

the x-axis◦ A line parallel to

the y-axis

You have …◦ the point (0,0)◦ the point (0,y)◦ the point (x,0)◦ the slope m = 0 eqn is y = ____◦ The slope m

undefined eqn is x = ____

Parallel & PerpendicularParallel & Perpendicular II II | |

Parallel slopes are equal

m original

m|| = mo

Perpendicular slopes are opposite reciprocals

m original

m | = -1 / mo

Linear equation partsLinear equation partsSlope

InterceptStandard Horizontal Vertical

Equationy = mx +

bAx + By =

C y = b x = a

Slope m-A

B0 undefined

y – intercept

bC

Bb

does not exist

x - intercept

-b m

C A

does not exist

a

parallel slope ||

m-A

B0 undefined

perpendicular slope _|_

-1 m

B A

undefined 0

Given 3x + 4y = 12Given 3x + 4y = 12Find the line || to

the given line that passes through (-2, 5).

Find the line _|_ to the given line that passes through (-2, 5).

Given 3x + 4y = 12Given 3x + 4y = 12Find the line || to

the given line that passes through (-2, 5).

Find the line _|_ to the given line that passes through (-2, 5).

If the line is parallel then the slope must be the same so the linear

equation will look like 3x + 4y =

If the line is perpendicular then the slope must be the opposite

reciprocal so the linear equation will look like -4x + 3y =

Given 3x + 4y = 12Given 3x + 4y = 12Find the line || to

the given line that passes through (-2, 5).

3x + 4y = ___3(-2) + 4(5) = ___-6 + 20 = 14

Find the line _|_ to the given line that passes through (-2, 5).

-4x + 3y = ___-4(-2)+3(5) = ___8+15 = 23

3x + 4y = 143x + 4y = 14 -4x + 3y = 23-4x + 3y = 23

Given y = 2x - 12Given y = 2x - 12Find the line || to

the given line that passes through (-2, 5).

Slope = 2 thereforem|| = 2

Find the line _|_ to the given line that passes through (-2, 5).

Slope = 2 thereforem_|_ = -1/2

y = 2x + by = 2x + b y = -1/2 x + by = -1/2 x + b

Given y = 2x - 12Given y = 2x - 12Find the line || to

the given line that passes through (-2, 5).

Find the line _|_ to the given line that passes through (-2, 5).

y = 2x + by = 2x + b5 = 2(-2) + b5 = 2(-2) + bb = 9b = 9

y = -1/2 x + by = -1/2 x + b5 = -1/2 (-2) + 5 = -1/2 (-2) + bbb = 4b = 4

y = -1/2 x + 4y = -1/2 x + 4y = 2x + 9y = 2x + 9

The alternative calculation is The alternative calculation is to using the point slope form to using the point slope form of a linear equation y – yof a linear equation y – y11 = = m(x – xm(x – x11))Once you identify

the desired slope, you have m

then you can substitute the point value for (x1,y1)

y = -3x – 7y = -3x – 7parallel through

(1,2)

perpendicular through (1,2)

y – 2 = -3(x – 1) y – 2 = 1/3(x – 1)

Find (e,f)Find (e,f)

remember if you can find the remember if you can find the blue lineblue line

you can find the y - you can find the y - interceptinterceptthen consider the then consider the reflectionreflection

Find tFind t

Select t so that the triangle with vertices ( -4, 2 ), ( 5, 1 ), and (t,-1) a right triangle with the right angle at (t,-1).

Find tFind t

Select t so that the triangle with vertices ( -4, 2 ), ( 5, 1 ), and (t,-1) a right triangle with the right angle at (t,-1).

Right angle

Switching gears…Switching gears…Parametric

equations of the line p. 69…