Cornell Notes

Learning Target (LT):I can identify and use parallel and perpendicular lines.(G.GPE.5)

Name:Class: Coord. AlgebraPeriod:Date: March 17, 2014

Questions Notes

WARM-UP:

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Cornell Notes

Learning Target (LT):I can identify and use parallel and perpendicular lines.(G.GPE.5)

Name:Class: Coord. AlgebraPeriod:Date: March 17, 2014

Questions

Notes

Parallel lines: Lie in the same plane BUT never intersect. On a coordinate plane, parallel lines have the same slope BUT

difference y-intercepts

Consider . . .

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Cornell Notes

Learning Target (LT):I can identify and use parallel and perpendicular lines.(G.GPE.5)

Name:Class: Coord. AlgebraPeriod:Date: March 17, 2014

Questions

Notes

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Perpendicular lines: Lie in the same plane AND intersect to form right angles. On a coordinate plane, perpendicular lines have slopes that are opposite

reciprocals. If one line has the slope of “m”, then the line perpendicular to is will have a slope of −1m .

The product of the slopes for two perpendicular lines is always -1 (with only one exception, where m = 0).

m∙− 1m

=−1

Consider . . .

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Cornell Notes

Learning Target (LT):I can identify and use parallel and perpendicular lines.(G.GPE.5)

Name:Class: Coord. AlgebraPeriod:Date: March 17, 2014

Questions

Notes

Example A: Are lines “a” and “b” parallel? How do you know?

Step A. What do you see?

Step B: What are the slopes of the lines, AND compare.

Step C. What are the y-intercepts of “a” and “b”

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Cornell Notes

Learning Target (LT):I can identify and use parallel and perpendicular lines.(G.GPE.5)

Name:Class: Coord. AlgebraPeriod:Date: March 17, 2014

Questions

Notes

Example B: The equations of two lines are: Line r: 4y + 12 = x; Line s: 8y + 2x = 16.

Are lines r & s perpendicular?

Step 1: Find the slopes of the two lines:

Let’s re-write the equation of the line into slope-intercept form. THE COEFFICIENT OF “x” WILL BE OUR SLOPE.

Line r Line s

Step 2: Compare the slopes of “r” and “s”.

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Cornell Notes

Learning Target (LT):I can identify and use parallel and perpendicular lines.(G.GPE.5)

Name:Class: Coord. AlgebraPeriod:Date: March 17, 2014

Questions

Notes

Example C: The equations of two lines are: Line t: 2y + 2 = 6x; Line u: A line parallel to line t, passing through the point (6,2)Line v: A line perpendicular to line t, passing through point (6,2)

What are the equations for lines u and v?

Step 1: Find the slope of line t since this line is connected to both u and v.

Step 2: Find the equation for line u.

Step 3: Find the equation for line v.

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Key Learnings:1. 2. 3.

Cornell Notes

Learning Target (LT):I can identify and use parallel and perpendicular lines.(G.GPE.5)

Name: TEACHER NOTESClass: Coord. AlgebraPeriod:Date: March 17, 2014

Questions Notes

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WARM-UP:

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Cornell Notes

Learning Target (LT):I can identify and use parallel and perpendicular lines.(G.GPE.5)

Name: TEACHER NOTESClass: Coord. AlgebraPeriod:Date: March 17, 2014

Questions

Notes

Parallel lines: Lie in the same plane BUT never intersect. On a coordinate plane, parallel lines have the same slope BUT

difference y-intercepts

Consider . . .

10

Cornell Notes

Learning Target (LT):I can identify and use parallel and perpendicular lines.(G.GPE.5)

Name: TEACHER NOTESClass: Coord. AlgebraPeriod:Date: March 17, 2014

Questions

Notes

11

Perpendicular lines: Lie in the same plane AND intersect to form right angles. On a coordinate plane, perpendicular lines have slopes that are opposite

reciprocals. If one line has the slope of “m”, then the line perpendicular to is will have a slope of −1m .

The product of the slopes for two perpendicular lines is always -1 (with only one exception, where m = 0).

m∙− 1m

=−1

Consider . . .

12

Cornell Notes

Learning Target (LT):I can identify and use parallel and perpendicular lines.(G.GPE.5)

Name: TEACHER NOTESClass: Coord. AlgebraPeriod:Date: March 17, 2014

Questions

Notes

Example A: Are lines “a” and “b” parallel? How do you know?

Step A. What do you see?

The lines do not intersect at any visible points, so they COULD BE parallel.

HOWEVER, the lines may intercept of the visual grid, so observation is NOT ENOUGH.

Discussion: Still a visual can be useful. Why?

Step B: What are the slopes of the lines, AND compare.

Find two points on line “a” (say points (0,4) and (-2, 1)). Calculate the slope using our point formula: y2− y1

x2−x1= 4−10−(−2)

= 32.

Find two points on line “b” (say points (0,-2) and (2, 1)). Calculate the slope using our point formula:

y2− y1x2−x1

=−2−10−2

=−3−2

=32.

Compare: slope of “a” is 3/2 and slope of “b” is 3/2. Are they parallel? Could they be collinear? We need one more piece of information . . . .

Step C. What are the y-intercepts of “a” and “b”

For “a”, the y-intercept is (0,4). For “b”, the y-intercept is (0,-2). Because the y-intercepts are DIFFERENT, the lines are NOT collinear.

Recapping: same slope AND different y-intercepts they are parallel.

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Cornell Notes

Learning Target (LT):I can identify and use parallel and perpendicular lines.(G.GPE.5)

Name: TEACHER NOTESClass: Coord. AlgebraPeriod:Date: March 17, 2014

Questions

Notes

Example B: The equations of two lines are: Line r: 4y + 12 = x; Line s: 8y + 2x = 16.

Are lines r & s perpendicular?

Step 1: Find the slopes of the two lines:

Let’s re-write the equation of the line into slope-intercept form. THE COEFFICIENT OF “x” WILL BE OUR SLOPE.

Line r Line s4y + 12 = x4y = x – 12 Y = ¼x - 3

8y + 2x = 168y = -2x + 16 Y = -¼x + 2

Step 2: Compare the slopes of “r” and “s”.

(Slope r) x (slope s) = 14 ∙−14=−116≠−1.

Since the product of the slopes -1, the lines are not perpendicular.

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Cornell Notes

Learning Target (LT):I can identify and use parallel and perpendicular lines.(G.GPE.5)

Name: TEACHER NOTESClass: Coord. AlgebraPeriod:Date: March 17, 2014

Questions

Notes

Example C: The equations of two lines are: Line t: 2y + 2 = 6x; Line u: A line parallel to line t, passing through the point (6,2)Line v: A line perpendicular to line t, passing through point (6,2)

What are the equations for lines u and v?

Step 1: Find the slope of line t since this line is connected to both u and v.

Let’s re-write the equation of the line into slope-intercept form. THE COEFFICIENT OF “x” WILL BE OUR SLOPE.

2y + 2 = 6x2y = 6x – 2 y = 3x - 1 . . . so slope of t = 3

Step 2: Find the equation for line u.

Because line u ∥ line t, the slope of u = slope of t = 3We also know that line u passes through (6,2)Using our point-slope equation y− y1=m (x−x1 )So . . . y – 2 = 3(x – 6) . . . which converts to the slope-intercept form y = 3x - 16

Step 3: Find the equation for line v.

Because line v ⊥ line t, the slope of v = negative reciprocal of 3, or −13

We also know that line u passes through (6,2)Using our point-slope equation y− y1=m (x−x1 )

So . . . y – 2 = −13 (x – 6) . . .

which converts to the slope-intercept form y = −13 x + 4

Key Learnings:1. Parallel lines have identical slopes AND different y-intercepts2. Perpendicular lines have slopes that are “negative reciprocals3. Each questions requeires a plan to solve the problem.

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