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Essential Question: How can you use the equations of two non-vertical lines to tell whether the equations are parallel or perpendicular?
Graphing a Linear Equation Any equation with both x and y can be graphed. A SOLUTION OF A LINEAR EQUATION IS ANY
ORDERED PAIR (x, y) THAT MAKES THE EQUATION TRUE.▪ In the equation
y = 3x + 2if x = 4then y = 3(4) + 2 = 14
▪ Which means (4, 14) is a solution to that equation.▪ There are infinite solutions to a linear equation, you just
need to decide on an x value to start with.▪ BECAUSE y DEPENDS ON THE VALUE OF x, WE CALL y
THE DEPENDENT VARIABLE, AND CALL x THE INDEPENDENT VARIABLE
Graphing a Linear Equation You can graph a linear equation by taking two
solutions, putting them on a coordinate plane, and connecting a line through them.
Example: Graph the equation y = 2/3x + 3
Graphing a Linear Equation You can graph a linear equation by taking two
solutions, putting them on a coordinate plane, and connecting a line through them.
Example: Graph the equation y = 2/3x + 3▪ Let x = 3, then y = 2/3(3) + 3 = 5
▪ Use the point (3, 5)
Graphing a Linear Equation You can graph a linear equation by taking two
solutions, putting them on a coordinate plane, and connecting a line through them.
Example: Graph the equation y = 2/3x + 3▪ Let x = 3, then y = 2/3(3) + 3 = 5
▪ Use the point (3, 5)
▪ Let x = -3, then y = 2/3(-3) + 3 = 1
▪ Use the point (-3,1)
Graphing a Linear Equation Use the point (3, 5) Use the point (-3,1)
(3, 5)
(–3, 1)
1 2 3 4 5–1–2–3–4–5 x
1
2
3
4
5
–1
–2
–3
–4
–5
y
Connect the line
The y-intercept is where the graph crosses the y-axis. It can be found by setting x = 0 in a linear
equation. The x-intercept is where the graph crosses
the x-axis. It can be found by setting y = 0 in a linear
equation. The intercepts can also be used in
graphing a linear equation.
Example: The equation3x + 2y = 120
models the number of passengers who can sit in a train car, where
x is the number of adults and y is the number of children.
Graph the equation. Explain what the x- and y-intercepts represent.Describe the domain and range.
3x + 2y = 120 x-intercept▪ 3x + 2(0) = 120▪ 3x = 120▪ x = 40
3x + 2y = 120 x-intercept▪ 3x + 2(0) = 120▪ 3x = 120▪ x = 40
y-intercept▪ 3(0) + 2y = 120▪ 2y = 120▪ y = 60
3x + 2y = 120 x-intercept▪ 3x + 2(0) = 120▪ 3x = 120▪ x = 40
y-intercept▪ 3(0) + 2y = 120▪ 2y = 120▪ y = 60
Use the points (40,0) and (0,60)
(0, 60)
(40, 0)
5 10 15 20 25 30 35 40 45 50 55 60 x
5
10
15
20
25
30
35
40
45
50
55
60
y
Slope Slope is found by taking the vertical change
and dividing by the horizontal change Rise over Run The formula is:
where (x1,y1) and (x2,y2) represent solutions to a linear equation
2 1
2 1
y y
x x
Slope Example: Find the slope of the line
through the points (3,2) and (-9,6)
12
12
2 4 1
12 33
6
9x
y y
x
Writing Equations of Lines Slope-Intercept Form▪ Used when you know the slope and the
y-intercept▪ y = mx + b
What does the m stand for? What does the b stand for?
Example: Find the slope of 4x + 3y = 7
Writing Equations of Lines Slope-Intercept Form▪ Used when you know the slope and the
y-intercept▪ y = mx + b
slope y-intercept
▪ Once you get y by itself, the slope is the coefficient in front of the “x”.
Example: Find the slope of 4x + 3y = 7
4x + 3y = 7 Get y by itself
4x + 3y = 7-4x -4x
3y = -4x + 7
4x + 3y = 7-4x -4x
3y = -4x + 73 3 3
y = -4/3 x + 7/3 (leave as fractions)
4x + 3y = 7-4x -4x
3y = -4x + 73 3 3
y = -4/3 x + 7/3 (leave as fractions)
Slope = -4/3
4x + 3y = 7-4x -4x
3y = -4x + 73 3 3
y = -4/3 x + 7/3 (leave as fractions)
Slope = -4/3
y-intercept = 7/3
Assignment Page 67 1 – 19, 33 – 37 (odd problems) SHOW WORK
Essential Question: How can you use the equations of two non-vertical lines to tell whether the equations are parallel or perpendicular?
Writing Equations of Lines Point-Slope Form▪ Used when you know a point on the line and
the slope
▪ y – y1 = m(x – x1)
Example: Write in slope-intercept form an equation of the line with slope -1/2 through the point (8,-1)
Slope = -1/2 Point = (8, -1)y – y1 = m(x – x1)
Slope = -1/2 Point = (8, -1)y – y1 = m(x – x1) (replace)y – (-1) = -1/2(x – 8)
Slope = -1/2 Point = (8, -1)y – y1 = m(x – x1) (replace)y – (-1) = -1/2(x – 8) (distribute)y + 1 = -1/2x + 4
Slope = -1/2 Point = (8, -1)y – y1 = m(x – x1) (replace)y – (-1) = -1/2(x – 8) (distribute)y + 1 = -1/2x + 4 (subtract 1) -1 -1y = -1/2 x + 3
Writing Equations of Lines Sometimes, you will be given two points.
In this case, you will first need to find the slope of the two points, then use either one of the points and the slope in point-slope form
Example: Write in point-slope form an equation of the line through (1,5) and (4,-1)
Writing Equations of Lines Sometimes, you will be given two points.
In this case, you will first need to find the slope of the two points, then use either one of the points and the slope in point-slope form
Example: Write in point-slope form an equation of the line through (1,5) and (4,-1)
Find the slope: 2 1
2 1
1 5 62
4 1 3
y ym
x x
Writing Equations of Lines Example: Write in point-slope form an
equation of the line through (1,5) and (4,-1)
Find the slope:
Choose either point▪ y – y1 = m(x – x1)
▪ y – 5 = -2(x – 1) OR y + 1 = -2(x – 4)will give valid answers
2 1
2 1
1 5 62
4 1 3
y ym
x x
The slopes of horizontal, vertical, parallel and perpendicular lines have special properties
Horizontal Line
Vertical Line Parallel Lines
Perpendicular Lines
m = 0 m = undefined Slopes are equal
Slopes are inverse reciprocals
y is constant x is constant Flip the fractionFlip the sign
Write an equation of the line through the point (6,1) and perpendicular to y = ¾ x + 2
What is the slope of my starting line?
Write an equation of the line through the point (6,1) and perpendicular to y = ¾ x + 2
What is the slope of my starting line? ¾ What is the slope of my perpendicular line?
Write an equation of the line through the point (6,1) and perpendicular to y = ¾ x + 2
What is the slope of my starting line? ¾ What is the slope of my perpendicular line?
Flip fraction = 4/3
Flip sign = -4/3
Leave your answer in slope-intercept form You have the slope, so find the intercept Note: You can also solve using point-slope form
Slope = -4/3 Point = (6, 1)y = mx + b
Find b 1 = -4/3 (6) + b (replace) 1 = -8 + b +8 +8 9 = b
y = -4/3 x + 9
Assignment Page 68 #21,23, 25 (Write in slope-intercept form) #27 – 31 (odd problems) #38 & 39