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Linear Equations Review Chapter 5 Chapters 1 & 2.

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Presentation on theme: "Linear Equations Review Chapter 5 Chapters 1 & 2."— Presentation transcript:

1 Linear Equations Review Chapter 5 Chapters 1 & 2

2 What you should know about Linear equations Slope Y-intercept X-intercept What does the graph look like? Parallel slope Perpendicular slope Given ANY linear equation you should be able to identify…

3 Equation Forms Slope Intercept Standard Horizontal Vertical y = mx + b Ax + By = C y = b x = a

4 Slopes NegativePositive Horizontal Vertical

5 Can you run through the linear equation information… 3x+4y=24 3x+4y=24 y = 1/2x-7 y = 1/2x-7 y = 5 y = 5 x = 6 x = 6

6 Given any linear equation, one should be able to 3x+4y=24 The Equation Form Direction Slope y-intercept x-intercept Parallel Slope Perpendicular Slope 1. Standard 2. Falling 3. -3/4 4. 6 5. 8 6. -3/4 7. 4/3 identify…

7 Given any linear equation, one should be able to y = 1/2x-7 The Equation Form Direction Slope y-intercept x-intercept Parallel Slope Perpendicular Slope 1. Slope intercept 2. Rising 3. 1/2 4. -7 5. - -7/(1/2) = 14 6. 1/2 7. -2 identify…

8 Given any linear equation, one should be able to y = 5 The Equation Form Direction Slope y-intercept x-intercept Parallel Slope Perpendicular Slope 1. Horizontal line 2. horizontal 3. 0 4. 5 5. Does not exist 6. 0 7. undefined identify…

9 Given any linear equation, one should be able to x = 6 The Equation Form Direction Slope y-intercept x-intercept Parallel Slope Perpendicular Slope 1. Vertical line 2. verticle 3. undefined 4. Does not exist 5. 6 6. undefined 7. 0 identify…

10 To 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”

11 (0,-7) (-7/3,0) Graph using intercepts y = -3x – 7 y-int = -7 x-int = 7 -3

12 up 3 back 1 (0,-7) down 3 over 1 Graph using intercepts y = -3x – 7 point= (0, -7) slope = -3 / 1

13 The slope formula m = y 1 – y 2 x 1 – x 2 This is really the same at the point-slope equation m(x 1 – x 2 ) = y 1 – y 2 y 1 – y 2 = m(x 1 – x 2 )

14 Find the slope given 2 points (-1,1); (2,3) m = 3 – 1 2 – -1 m = 2 3 m = 1 – 3 -1 – 2 m = -2 = 2 -3 3

15 Now write the equation (-1,1); (2,3) m = 3 – 1 2 – -1 m = 2 3 y 1 – y 2 = m(x 1 – x 2 ) y – 3 = 2/3 (x – 2) y = 2/3 x – 4/3 + 9/3 y = 2/3 x + 5/3

16 If you have two points you can find the line…sometimes the challenge is 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 = ____

17 Parallel & Perpendicular II | Parallel slopes are equal m original m || = m o Perpendicular slopes are opposite reciprocals m original m | = -1 / m o

18 Linear equation parts Slope Intercept StandardHorizontalVertical Equation y = mx + bAx + By = C y = bx = a Slopem -A B 0undefined y – interceptb CBCB bdoes not exist x - intercept -b m CACA does not exista parallel slope || m -A B 0undefined perpendicular slope _|_ -1 m BABA undefined0

19 Given 3x + 4y = 12 Find the line || to the given line that passes through (-2, 5). Find the line _|_ to the given line that passes through (-2, 5).

20 Given 3x + 4y = 12 Find 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 = 

21 Given 3x + 4y = 12 Find 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 = 14 -4x + 3y = 23

22 Given y = 2x - 12 Find the line || to the given line that passes through (-2, 5). Slope = 2 therefore m || = 2 Find the line _|_ to the given line that passes through (-2, 5). Slope = 2 therefore m _|_ = -1/2 y = 2x + b y = -1/2 x + b

23 Given y = 2x - 12 Find 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 + b 5 = 2(-2) + b b = 9 y = -1/2 x + b 5 = -1/2 (-2) + b b = 4 y = -1/2 x + 4 y = 2x + 9

24 The alternative calculation is to using the point slope form of a linear equation y – y 1 = m(x – x 1 ) Once you identify the desired slope, you have m then you can substitute the point value for (x 1,y 1 ) y = -3x – 7 parallel through (1,2) perpendicular through (1,2) y – 2 = -3(x – 1) y – 2 = 1/3(x – 1)

25 Find (e,f)

26 remember if you can find the blue line you can find the y - intercept then consider the reflection

27 Find 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).

28 Find 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

29 Switching gears… Parametric equations of the line p. 69…


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