Created by Charlean Mullikin: ML sections 3.6/3.7

Slope is the relationship of the rise to the run of a line. m = rise = y 2 – y 1 run x 2 – x 1

Slope can be positive: + ÷ + or - ÷ - Slope can be negative: + ÷ - or - ÷ +

Slope can be 0: 0 ÷ anything Slope can be undefined: Anything ÷ 0 Horizontal Vertical

ALWAYS SIMPLIFY SLOPES Slopes are positive, negative, 0, or Undefined (No slope). Slopes are written as integers with one sign, proper fractions, or improper fractions (no mixed fractions). When 0 is on top, the slope is 0. When 0 is on bottom, the slope is undefined or no slope. m = 5 m = -4 m = 1/3 m = - 3/5 m = 5/2 m = 0 m = undefined m = 5 1/2 m = -5/-3 m = 15/3 m = -15/-25m = 0/6 m = 5/0 m = 12/-8

m = rise = y 2 – y 1 run x 2 – x 1 (x 1, y 1 ) (x 2, y 2 )y2y2 x2x2 x1x1 y1y1 – – Rise On top!! Run On bottom!!

m = rise = y 2 – y 1 = run x 2 – x 1 (3, -3) (0, 9) – – Rise On top!! Run On bottom!! Find the slope of the line that passes through (3, -3)and (0, 9) -3 0 = – 9 3 – = - 4

a: m= +5 =1 b: m= +2 =1 YES, Since the slopes are the same (1=1), then the lines ARE PARALLEL.

6/2 = 3-10/-2 = 5-24/8 = -3 2/6 = 1/39/0 = undefined0/22 = 0

Application Identify rise and run. Which word points to the rise? Put the rise on top. What is the run? Put the run on bottom. Change to same units, then Divide out and Answer the question in reasonable units feet 3.1 miles = The average slope is about x 5280 = ft feet

Perpendicular Lines  When two lines are perpendicular, there are two cases with relation to slopes:  Case 1-If neither line is vertical, the product of the two slopes is negative one (Opposite reciprocals). m 1 =2/3 and m 2 = - 3/2  Case 2 – If one of the lines is vertical, then the perpendicular line is horizontal. m 1 =undefined and m 2 = 0

What is the slope of….. Slope of given lineParallel Line?Perpendicular Line? 1/2 -6 3/5 -8/7 0 4 No slope 1/2 -6 3/5 -8/7 0 4 No slope -2 1/6 -5/3 7/8 No slope -1/4 0

Writing Equations Shortcut #1 1

Writing Equations Shortcut #2 1

Writing Equations Writing Equations

Shortcut #1 Shortcut #2

Writing Equations Writing Equations

1 1

Identify ONE point to use Find slope Substitute

Simplify and solve for y Distributive Property of = Addition Property of = (Add 8 to both sides) Combine like terms Use calculator!

Parallel Equations LLLLines that are parallel have the same slope. –I–I–I–Identify slope of given line –I–I–I–Identify point parallel line passes through –U–U–U–Use point-slope equation to write equation

WWWWrite the equation of the line parallel to y = ¾ x – 5 that passes through the point (3, -2). mmmm = ¾, parallel slope is also ¾ PPPPoint (3, -2) yyyy – y1 = m(x – x1) yyyy - -2 = ¾(x – 3) yyyy + 2 = ¾ x – 9/4 yyyy = ¾ x – 9/4 – 2 yyyy = ¾ x – 17/4

WWWWrite the equation of the line parallel to 7x + 5y = 13 that passes through the point (1, 2). SSSSolve for y to find slope: 7777x + 5y = 13 5555y = -7x + 13 (subtract 7x from both sides) yyyy = -7/5 x + 13/5 (Divide each term by 5) pppparallel slope is – 7/5 PPPPoint (1, 2) yyyy – y1 = m(x – x1) yyyy - 2 = - 7/5 (x – 1) yyyy - 2 = -7/5 x + 7/5 yyyy = -7/5 x + 7/5 + 2 yyyy = -7/5 x + 17/5

Perpendicular Equations LLLLines that are perpendicular have slopes that multiply to equal -1. They are opposite sign, reciprocal numbers. –I–I–I–Identify slope of given line –C–C–C–Change the sign and flip the number to get the perpendicular slope. –U–U–U–Use point-slope equation to write equation

WWWWrite the equation of the line perpendicular to 7x + 5y = 13 that passes through the point (1, 2). SSSSolve for y to find slope: 7777x + 5y = 13 5555y = -7x + 13 yyyy = -7/5 x + 13/5 pppperpendicular slope is +5/7 PPPPoint (1, 2) yyyy – y1 = m (x – x1) yyyy - 2 = +5/7(x – 1) yyyy - 2 = 5/7 x – 5/7 yyyy = 5/7 x – 5/7 + 2 yyyy = 5/7 x + 9/7

WWWWrite the equation of the line perpendicular to y = ¾ x – 5 that passes through the point (3, -2). mmmm = ¾, perpendicular slope is – 4/3 PPPPoint (3, -2) yyyy – y1 = m(x – x1) yyyy - -2 = -4/3(x – 3) yyyy + 2 = -4/3 x + 4 yyyy = -4/3 x + 4 – 2 yyyy = -4/3 x + 2