1 PRIMARY CONTENT MODULE Algebra I -Linear Equations & Inequalities T-71 Applications The formula y = mx + b sometimes appears with different symbols. For example, instead of x, we could use the letter C. Instead of y, we could use the letter F. Then the equation becomes F = mc + b. All temperature scales are related by linear equations. For example, the temperature in degrees Fahrenheit is a linear function of degrees Celsius.
2 PRIMARY CONTENT MODULE Algebra I -Linear Equations & Inequalities T-72 Basic Temperature Facts Water freezes at: 0 C, 32 F Water Boils at: 100 C, 212 F F (100, 212) (0, 32) C Can you solve for m and b in F = mc + b?
3 PRIMARY CONTENT MODULE Algebra I -Linear Equations & Inequalities T-73 To find the equation relating Fahrenheit to Celsius we need m and b m = F 2 F 1 C 2 C m = m = = 9 5 Therefore F = 9 5 C + b To find b, substitute the coordinates of either point. Therefore b = = 9 5 (0) + b Therefore the equation is F = 9 5 C + 32 Can you solve for C in terms of F?
4 PRIMARY CONTENT MODULE Algebra I -Linear Equations & Inequalities T-74 Celsius in terms of Fahrenheit F = 9 5 C F = C F = C (32) 5 9 F 5 9 (32) = C C = 5 9 F 5 9 (32) C = 5 (F 32) 9 Example: How many degrees Celsius is77 F? C = 5 (77 32) 9 C = 5 9 (45) C = 25 So 72 F = 25 C
5 PRIMARY CONTENT MODULE Algebra I -Linear Equations & Inequalities T-75 Standard 8 Algebra I, Grade 8 Standards Students understand the concept of parallel and perpendicular lines and how their slopes are related. Students are able to find the equation of a line perpendicular to a given line that passes through a given point.
6 PRIMARY CONTENT MODULE Algebra I -Linear Equations & Inequalities T-76/H-76 On one clean sheet of graph paper: 1. Graph y = 2x 2. Graph y = 2x + 3
7 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-77 Solution: y = 2x + 3 slope = 2; y-intercept = 3 y = 2x slope = 2; y-intercept = 0
8 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-78 Notice that y = 2x and y = 2x + 3 are parallel. If two lines have the same slope and different y-intercepts, they are parallel. Remember that a different y-intercept only translates a graph vertically, so the translated graph is parallel to the original.
9 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-79 Theorems Two non-vertical lines are parallel if and only if they have the same slope. Two non-vertical lines are perpendicular if and only if the product of their slopes is 1.
10 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities H-80 Practice Consider the line whose equation is y = 2x 5. Questions: 1. If a line is parallel to the graph of the given line, what is its slope? 2. If a line is perpendicular to the graph of the line, what will be its slope?
11 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-80 Practice Consider the line whose equation is y = 2x 5. Questions: 1. If a line is parallel to the graph of the given line, what is its slope? 2. If a line is perpendicular to the graph of the line, what will be its slope? Answers: 1. m = 2 2. m = 1 2
12 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-81 More Practice Find the equation of the line whose graph contains the point (2,1) and is perpendicular to the line with the equation y = 2x 5. y = 2x 5 (2, 1) 5 y =?
13 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-82 Perpendicular Lines Write the equation of the line that passes through the point (2,1) and is perpendicular to the line whose equation is y = 2x 5. The slope of the line is 2, so the slope of the line that is perpendicular is 1 2. Point-Slope Formula Slope-Intercept Formula y y 1 = m(x x 1 ) y = mx + b y 1 = 1 2 (x 2) 1 = 1 2 (2) + b y 1 = 1 (x 2) 1 = 1 + b 2 y 1 = x = b y = 1 2 x + 2 y = 1 2 x + 2
14 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-83/H-83 Practice Write the equation of the line that passes through the point (5,1) and is perpendicular to the line whose equation is y = 1 3 x + 4.
15 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-84 Solution The equation of the line y = 1 3 x + 4 has slope of 1 3 so the slope of a line that is perpendicular is 3. Our point was (5,1). Point-Slope Formula Slope-Intercept Formula y y 1 = m(x x 1 ) y = mx + b y 1 = 3(x 5) 1 = 3(5) + b y 1 = 3x = 15 + b y = 3x = b; b = 16 y = 3x + 16
16 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-85 In-Depth Understanding Why is it that two non-vertical lines are perpendicular if and only if the product of their slopes is 1? The explanation uses the Pythagorean Theorem and its converse. What is the Pythagorean Theorem?
17 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-86 PYTHAGOREAN THEOREM b c a 2 + b 2 = c 2 Example: Find the missing length. x a 5 12 x 2 = x 2 = x 2 = 169; x = 13
18 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-87 An Important Tool: The Distance Formula The distance between two points (x 1, y 1 ) and (x 2, y 2 ) is given by the distance formula d = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 d (x 2, y 2 ) (x 1, y 1 ) This comes from the Pythagorean Theorem.
19 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-88 The Distance Formula (x 2, y 2 ) (d) (y 2 y 1 ) (x 1, y 1 ) (x 2 x 1 ) From Pythagoras Theorem d 2 = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 d = (x 2 x 1 ) 2 + (y 2 y 1 ) 2
20 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-89 Determining the Length of a Segment So for example given (2,5) and (3,9) the distance between these points is d = (2 3) 2 +(5 9) 2 d = ( 1) 2 + ( 4) 2 d = 1+16 d = 17
21 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-90 We will need the converse of the Pythagorean Theorem to understand why lines whose slopes have a product of 1 must necessarily be perpendicular. Converse of the Pythagorean Theorem: If the lengths of the sides of a triangle, a, b, and c, satisfy a 2 + b 2 = c 2, then the triangle is a right triangle.
22 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities H-91 One More Piece of the Puzzle A little algebra: (a + b) 2 = (a + b)(a + b) = A(a + b) (think of A = (a + b)) = Aa + Ab = (a + b)a + (a + b)b = a 2 + ba + ab + b 2 = a 2 + ab + ab + b 2 = a 2 + 2ab + b 2 Now try: (a c) 2 = (b d) 2 =
23 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-91 One More Piece of the Puzzle A little algebra: (a + b) 2 = (a + b)(a + b) = A(a + b) (think of A = (a + b)) = Aa + Ab = (a + b)a + (a + b)b = a 2 + ba + ab + b 2 = a 2 + ab + ab + b 2 Now try: = a 2 + 2ab + b 2 (a c) 2 = (b d) 2 = Answer: (a c) 2 = a 2 2ac + c 2 (b d) 2 = b 2 2bd + d 2
24 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-92 Goal: Show that two non-vertical lines are perpendicular if and only if the product of their slopes is 1. For simplicity, assume that both intersecting lines have y-intercepts equal to zero. y = m 1 x y = m 2 x
25 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-93 y = m 1 x C(c,d) A(a,b) B(0,0) Let (a,b) be a point on the graph of y = m 1 x and (c,d) be a point on the graph of y = m 2 x. Then m 1 = b 0 a 0 = b a and m 2 = d 0 c 0 = d c. Assume that triangle ABC is a right triangle. We want to show that m 1 m 2 = b a d c = 1. The Pythagorean Theorem says (AB) 2 + (BC) 2 = (AC) 2 AB = a 2 + b 2 AC = (a c) 2 + (b d) 2 BC = c 2 + d 2 y = m 2 x Now substitute
26 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-94 Perpendicular Lines Continued ( a 2 + b 2 ) 2 + ( c 2 + d 2 ) 2 = ( (a c) 2 +(b d) 2 ) 2 Remember our goal is to show that b m 1 m 2 = a d c = 1 a 2 + b 2 + c 2 + d 2 = (a c) 2 + (b d) 2 a 2 + b 2 + c 2 + d 2 = a 2 2ac + c 2 + b 2 2bd + d 2 Now subtract a 2, b 2, c 2, and d 2 from both sides. 0 = 2ac 2bd 2bd = 2ac bd = ac bd ac = 1 b a d c = 1 m 1 m 2 = 1 (Whew!!)
27 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-95 What have we done? We have shown that if two lines are perpendicular, then the product of their slopes is 1. Now suppose the product of the slopes is 1. Can we deduce that the lines are perpendicular? Yes! Start at the bottom of the previous slide and work backward. The last step is to apply the converse of the Pythagorean Theorem. Theorem: Two non-vertical lines are perpendicular if and only if the product of their slope is 1.
28 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-96 SAT Problem x-scale y-scale On the linear scales above, 20 and 60 on the x- scale correspond to 50 and 100, respectively, on the y-scale. Which of the following linear equations could be used to convert an x-scale value to a y-scale value? A) y = x + 30 B) y = x + 40 C) y = 1.25x + 25 D) y = 1.25x + 30 E) y = 0.8x + 34
29 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-97 Solving Inequalities Solving inequalities is similar to solving equations, except that instead of an equal sign one has an inequality. The main fact that you need to remember is that when you multiply or divide both sides of the inequality by a negative number, it reverses the inequality. Let s examine this number sentence 3 > 5 If I multiply both sides by 1 3( 1)? ( 5)( 1) 3 < 5 You can see that if we put the original inequality (greater than) in the blank, the sentence is no longer true. The inequality sign must change to less than (<). How do we graph linear inequalities?
30 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-98 Example Graph y > x + 2 First: Identify the equation of the line y = x + 2. x-intercept: When y = 0 what is x? 2 So coordinates of one point ( 2,0) y-intercept: When x = 0 what is y? 2 So coordinates of second point (0,2)
31 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-99 (The graph does not include the line y = x + 2) For a linear inequality of the form y>mx+b, the y-values satisfying the inequality are greater than the corresponding y-values which satisfy the equation y=mx+b. It follows that the halfplane above the graph is shaded. NOTE: The equation needs to be in slopeintercept form for the above assumption to be made. Also the coefficient of y should be +1.
32 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-100 What would the graph of 2x + 6y > 4 look like? How would the graph change if the problem was 2x + 6y 4?
33 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-101 From the inequality 2x + by > 4, we can isolate y to get y > 1 3 x + 2. The graph of this 3 inequality is the set of all points of the form (x,y) where y > 1 3 x If the problem were instead, 2x + 6y 4, then the graph would include the line.
34 PRIMARY CONTENT MODULE Algebra I - Linear Equations & Inequalities T-102 Graph the inequality: First isolate y: 2x 3y 9 3y 2x + 9 2x 3y 9 y 2 x 3 (Why did switch to?) 3
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