parallel and perpendicular lines in the cartesian plane

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Parallel and Perpendicular Lines in the Cartesian Plane

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The slide show review slope intercept form and provides instruction for a constructivist activity for students to discover the relationship between the slopes of two parallel or two perpendicular lines.

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Page 1: Parallel and perpendicular lines in the cartesian plane

Parallel and Perpendicular Lines in the

Cartesian Plane

Page 2: Parallel and perpendicular lines in the cartesian plane

They are boring!

They have no use in life.

STEREOTYPES ABOUT PARALLEL AND PERPENDICULAR LINES

Page 3: Parallel and perpendicular lines in the cartesian plane

Just a series of lines with positive slopes…

No Big Deal

Page 4: Parallel and perpendicular lines in the cartesian plane

Color coded to show

parallel and

perpendicular lines

Page 5: Parallel and perpendicular lines in the cartesian plane

WHOA!

Page 6: Parallel and perpendicular lines in the cartesian plane

I know… I’m Awesome!

Page 7: Parallel and perpendicular lines in the cartesian plane

PARALLEL AND PERPENDICULAR LINES ARE EVERYWHERE

Construction

Maps

Sports

Artwork

Page 8: Parallel and perpendicular lines in the cartesian plane

y = mx + b m is the slope of the lineb is the y-intercept

REVIEW: SLOPE INTERCEPT FORM

Life is easy when you’re

in slope

intercept form

Page 9: Parallel and perpendicular lines in the cartesian plane

y = mx + b

The y-intercept is the y value when x = 0.

Visually, the y-intercept is y value when the line crosses the y axis

http://www.mathsisfun.com/data/function-grapher.php

Y -INTERCEPT

Page 10: Parallel and perpendicular lines in the cartesian plane

m

y = mx + bSlope Slider

Slope ofvertical lines?

SLOPE

(𝑥1 , 𝑦1)

(𝑥2 , 𝑦 2)

Page 11: Parallel and perpendicular lines in the cartesian plane

3y = 6x + 9

5y = 10x

y = -1

x = 3

IDENTIFYING THE SLOPE AND THE Y-INTERCEPT

Hint

Page 12: Parallel and perpendicular lines in the cartesian plane

y = mx + bGiven the slope, m, and a point, (x , y),

then we can find b, the y-intercept.

b = y – mxOnce we find b, we can find the equation of the

line.

REVIEW: FINDING THE EQUATION OF THE LINE GIVEN A SLOPE AND A POINT ON THE LINE

Page 13: Parallel and perpendicular lines in the cartesian plane

p = (-2 , 2) m = 4

p = (-3 , 4) m = -2

p = (-2 , 2/3) m = -4/3

PRACTICE: FINDING THE EQUATION OF THE LINE GIVEN THE SLOPE AND A POINT ON THE LINE

Page 14: Parallel and perpendicular lines in the cartesian plane

1. Graph line segments. Be sure that each endpoint is an integer coordinate, such as (1,3) or (-3,0)Compute and record their slope.

2. Then graph a parallel line to each of the three line segments. Compute and record the slopes of the parallel lines. Then delete the parallel lines.

3. Then graph a perpendicular line to each of the three line segments. Compute and record the slopes of the perpendicular lines.

GRAPHING ACTIVITY

Page 15: Parallel and perpendicular lines in the cartesian plane
Page 16: Parallel and perpendicular lines in the cartesian plane
Page 17: Parallel and perpendicular lines in the cartesian plane
Page 18: Parallel and perpendicular lines in the cartesian plane

PARALLEL LINES

Two lines are parallel

The lines never

intersect

Slopes are equal

Page 19: Parallel and perpendicular lines in the cartesian plane

y = (1/3)x + 2

y – 1 = 6x

2y = 5x + 3

4y = 8x

y = 6

x = -3

FIND THE SLOPE OF A PARALLEL LINE

Page 20: Parallel and perpendicular lines in the cartesian plane

PERPENDICULAR LINES

Two lines are perpendicular

The lines intersect at right angle

Slopes are negative

reciprocals

Page 21: Parallel and perpendicular lines in the cartesian plane

y = -3x – 2

y = (1/3)x + 2

y – 1 = 6x

2y = 5x + 3

y = 6

x = -3

FIND THE SLOPE OF A PERPENDICULAR LINE

Page 22: Parallel and perpendicular lines in the cartesian plane

y = (1/3)x + 2 , p = (2 , -3)

2y = 5x + 3 , p = (1/2 , 2/3)

y = 6 , p = (6 , 0)

x = -3 , p = (1 , 2)

FIND THE EQUATION OF THE PARALLEL LINE THAT PASSES THROUGH THE GIVEN POINT.

Page 23: Parallel and perpendicular lines in the cartesian plane

y = -3x – 2 , p = (-1 , 4)

4y = 8x , p = (1 , 1/3)

y = 6 , p = (6 , 0)

x = -3 , p = (1 , 2)

FIND THE EQUATION OF THE PERPENDICULAR LINE THAT PASSES

THROUGH THE GIVEN POINT.