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Linear Equations Equations of Lines. 7/2/2013 Linear Equations 2 Lines and Equations Point-Slope Form Given line L and point (x 1, y 1 ) on L Let (x,

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Presentation on theme: "Linear Equations Equations of Lines. 7/2/2013 Linear Equations 2 Lines and Equations Point-Slope Form Given line L and point (x 1, y 1 ) on L Let (x,"— Presentation transcript:

1 Linear Equations Equations of Lines

2 7/2/2013 Linear Equations 2 Lines and Equations Point-Slope Form Given line L and point (x 1, y 1 ) on L Let (x, y) be any other point on L Find slope y x   (x 1, y 1 ) (x, y) L ∆x = x – x 1 ∆y = y – y 1 m = ∆y ∆x

3 7/2/2013 Linear Equations 3 Lines and Equations Point-Slope Form y x   L ∆x = x – x 1 ∆y = y – y 1 m = ∆y ∆x = y – y 1 x – x 1 Point-Slope Form y – y 1 = m(x –x 1 ) (x 1, y 1 ) (x, y)

4 7/2/2013 Linear Equations 4 Slope-Intercept Form Consider a non-vertical line L Locate y-intercept Let (x, y) be any other point on the line Find slope m y x L Lines and Equations   (0, b) (x, y) ∆x = x – 0 ∆y = y – b

5 7/2/2013 Linear Equations 5 Slope-Intercept Form Lines and Equations y x L   (0, b) (x, y) ∆x = x – 0 ∆y = y – b Slope-Intercept Form m(x – 0) = y – b y = mx + b Solving for y, m = ∆y ∆x = y – b x – 0

6 7/2/2013 Linear Equations 6 Examples Point-Slope Form A line through (5, 10) with slope Find the equation: y x  4 5 (5,10) m = 4/5 = y – 10 x – 5 Question: Is this form unique ? m = 4 5

7 7/2/2013 Linear Equations 7 Examples Slope-Intercept Form A line with intercept (0, 4) and slope -3 Find the equation: y = mx + b = -3x + 4 y x  (0, 4) m = -3 Question: Is this form unique ?

8 7/2/2013 Linear Equations 8 Lines and Equations Standard Form Algebraic form not directly graph related Useful for systems of linear equations For constants A, B and C, with B ≠ 0 Ax + By = C Can we still find slope and intercepts ? Question:Is this form unique ?

9 7/2/2013 Linear Equations 9 Lines and Equations Standard Form – Slope Rewriting the equation: This has slope-intercept form with = mx + b y = x A B – + C B = m Slope = A B – Note: a ≠ A b ≠ B,

10 7/2/2013 Linear Equations 10 Standard Form Rewriting the equation: Lines and Equations = mx + b y = x A B – + C B Note: b ≠ B = m A B – So … C B b= = b m – – C B A B – ( ) = C A – Intercepts Fractions

11 7/2/2013 Linear Equations 11 Standard Form Rewriting the equation: Intercepts Lines and Equations y = x A B – + C B Note: a ≠ A y-intercept = (0, b) ( ) 0, C B = x-intercept ( ) = b m –, 0 = C A, 0 – Intercepts b ≠ B ( a, 0) =

12 7/2/2013 Linear Equations 12 Intercepts with Standard Form Consider equation 6x + 5y = 30 When x = 0, 5y = 30 y = 6 Vertical intercept is (0, 6) Standard Form Example x y (0, 6) 

13 7/2/2013 Linear Equations 13 Intercepts with Intercept Form Consider equation 6x + 5y = 30 When y = 0, 6x = 30 x = 5 Horizontal intercept is (5, 0) Standard Form Example x y (0, 6) (5, 0)  

14 7/2/2013 Linear Equations 14 Intercept Form Consider standard form equation Ax + By = C For C ≠ 0, this becomes Lines and Equations A C x + B C y = 1 OR intercept form = 1 () C A x + y () C B Fractions

15 7/2/2013 Linear Equations 15 Intercept Form Consider intercept form equation Lines and Equations x y   () C A xy C B ( ) + = 1 0, C B ) (, 0 C A ( ) Question: What if C = 0 ? = 1 a x + b y ( a, ) 0 = ( ) 0, b= Note: A ≠ a B ≠ b

16 7/2/2013 Linear Equations 16 Finding Intercepts with Intercept Form Consider equation 6x + 5y = 30 Vertical intercept is (0, 6) Horizontal intercept is (5, 0) Standard Form Example x y (0, 6) (5, 0)   x + 5y ( ) 30 1 ( ) = x 5 y 6 + = 1

17 7/2/2013 Linear Equations 17 Horizontal and Vertical Lines Horizontal Lines Form: y = k for some constant k  From standard form Ax + By = C when A = 0, B ≠ 0 Question: What is the line y = 0 called ? C B y = The x-axis !

18 7/2/2013 Linear Equations 18 Horizontal and Vertical Lines Horizontal Lines Example: y = 3 Pick any points (x 1, 3), (x 2, 3) Slope m is then Note: Zero slope is NOT the same as no slope x y y = 3  (x 1, 3) (x 2, 3)   x1x1 x2x2  x = x 2 – x 1  y = 0 m = yy xx 3 – 3 x 2 – x 1 = 0 =

19 7/2/2013 Linear Equations 19 Horizontal and Vertical Lines Vertical Lines Form: x = k for some constant k From standard form Ax + By = C when A ≠ 0, B = 0 Then Ax = C A C x = Question: What is the line x = 0 ? The y-axis !

20 7/2/2013 Linear Equations 20 x y Vertical Line Example: x = 4 Pick any points (4, y 1 ), (4, y 2 ) Slope m is then Horizontal and Vertical Lines x = 4   (4, y 1 ) (4, y 2 ) y1y1  x = 0 m = yy xx = 4 – 4 y 2 – y 1 = 0  y2y2 yy  Undefined ! Note: No slope is not the same as zero slope !

21 7/2/2013 Linear Equations 21 Parallel Lines Horizontal Lines Zero slope, always parallel Vertical Lines No slope, always parallel Other lines Lines with same slope, always parallel

22 7/2/2013 Linear Equations 22 x y Parallel Lines Parallel Lines Example Find the equation of the line through (4, 10) parallel to the line Slope of new line is Point-slope form is (4, 10) y = – (½)x + 6 OR y = – (½)x + 12  1 2 – y 1 2 – = x + 6 = y – 10 x – – = y – 10 (x – 4) 1 2 –

23 7/2/2013 Linear Equations 23 Step 1 Geometry gives us Perpendicular Lines in General x y L2L2 L1L1 m2m2 m1m1 a c1c1 c2c2 c 1 2 = a 2 + b 1 2 c 2 2 = a 2 + b 2 2 m1m1 = b1b1 a m2m2 = – b 2 a, b1b1 am1am1 = b2b2 – a m 2 = b 1 = a m 1 b 2 = – a m 2, c c 2 2 = b12 b12 + 2a2 + b22+ 2a2 + b22

24 7/2/2013 Linear Equations 24 Step 2 Perpendicular Lines in General x y L2L2 L1L1 m2m2 m1m1 a c1c1 c2c2 b1b1 b2b2 = – 2 a 2 m 1 m 2 c c 2 2 = ( b 1 + b 2 ) 2 b1 + b2b1 + b2 = b b 1 b 2 + b 2 2 c c 2 2 = b b22+ b a 2 2a22a2 = 2 ( a m 1 )(– a m 2 ) 1 = – m 1 m 2 2 b 1 b 2 = – 1 m2m2 m1m1 =

25 7/2/2013 Linear Equations 25 Perpendicular Lines Example Example Find the equation of the line through (2, 3) perpendicular to line y = – (⅓) x + 3 Slope of the given line is m 1 = – ⅓ Slope of the new line is x y (2, 3) y = – ( ⅓ )x + 3 = 3 1 m1m1 – m2m2 =  = – 1 ⅓ –

26 7/2/2013 Linear Equations 26 Perpendicular Lines Example Example Slope of the new line is y = 3x – 3 = = 3 m2m2 1 m1m1 – = – 1 ⅓ – Alternate point-slope form y – y 1 = m(x – x 1 ) y – 3 = 3(x – 2) Slope-intercept form y = mx – mx 1 + y 1 y = 3x – 3 x y (2, 3)  y = – ( ⅓ )x + 3

27 7/2/2013 Linear Equations 27 Variation Direct Variation A variable y varies directly as variable x if y = kx for some constant k  The constant k is called the constant of variation  K is also known as the constant of proportionality

28 7/2/2013 Linear Equations 28 Variation Direct Variation Example State sales tax t varies directly as the amount of sale s, i.e. t = ks For tax of $200 on a $12.50 sale, what is the constant of variation ? s t k = t s = = k =.0625 Question: Does this look like y = mx + b ?

29 7/2/2013 Linear Equations 29  Inverse Variation Variable y varies inversely as variable x if for constant of variation k k is also known as the constant of inverse proportionality Variation x y y = k x

30 7/2/2013 Linear Equations 30 Variation Inverse Variation Example At constant temperature the pressure P of a gas in a balloon is inversely proportional to its volume V so that V P P = k V

31 7/2/2013 Linear Equations 31 Think about it !

32 7/2/2013 Linear Equations 32 Step 1 Geometry gives us Perpendicular Lines in General x y L2L2 L1L1 m2m2 m1m1 a c1c1 c2c2 c 1 2 = a 2 + b 1 2 c 2 2 = a 2 + b 2 2 m1m1 = b1b1 a m2m2 = – b 2 a, b1b1 am1am1 = b2b2 – a m 2 = b 1 = a m 1 b 2 = – a m 2 c c 2 2 =, = a 2 + ( a m 1 ) 2 = a 2 + ( a m 2 ) 2 a 2 + ( a m 1 ) 2 + a 2 + ( a m 2 ) 2

33 7/2/2013 Linear Equations 33 Parallel and Perpendicular Lines Perpendicular Lines in General (continued) Thus Subtracting like terms gives Recalling that b 1 = a m 1 and b 2 = – a m 2 gives  Thus x y L2L2 L1L1 m2m2 m1m1 a b1b1 c1c1 b2b2 c2c2 am1am1 = – a m 2 = 2 b 1 b 2 = 2 a 2 2( a m 1 )(– a m 2 ) = 2 a 2 b b 1 b 2 + b 2 2 = b a 2 + b 2 2 – 2 a 2 m 1 m 2 = 2 a 2 m 1 m 2 = –1 OR m 2 = 1 m1m1 –

34 7/2/2013 Linear Equations 34 Parallel and Perpendicular Lines Perpendicular Lines in General (continued) Thus Subtracting like terms gives Recalling that b 1 = a m 1 and b 2 = – a m 2 gives  Thus x y L2L2 L1L1 m2m2 m1m1 a b1b1 c1c1 b2b2 c2c2 am1am1 = – a m 2 = 2 b 1 b 2 = 2 a 2 2( a m 1 )(– a m 2 ) = 2 a 2 b b 1 b 2 + b 2 2 = b a 2 + b 2 2 – 2 a 2 m 1 m 2 = 2 a 2 m 1 m 2 = –1 OR m 2 = 1 m1m1 –


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