Presentation is loading. Please wait.

Presentation is loading. Please wait.

Linear Algebra Problem 3.3 Friday, September 5. Problem 3.2 answers.

Similar presentations


Presentation on theme: "Linear Algebra Problem 3.3 Friday, September 5. Problem 3.2 answers."— Presentation transcript:

1 Linear Algebra Problem 3.3 Friday, September 5

2 Problem 3.2 answers

3 Problem 3.2 ACE answers #4, #5

4 Problem 3.2 ACE answers #9

5 Learning Target I will be able to identify the coordinate rules that describe rotation for turns of 90 o and 180 o.

6 Problem 3.3 A Rotating 90 o counterclockwise

7 Problem 3.3 A Rotating 90 o counterclockwise answer for A 1 PointABCDE Original Coordinates(0,0)(2,4)(5,4)(6,6)(3,6) Coordinates after 90 o turn (0,0)(-4,2)(-4,5)(-6,6)(-6,3) Rule for rotating counterclockwise 90 o (x,y)  (-y,x)

8 Problem 3.3 A Rotating 90 o counterclockwise answer for A2 PointABCDE Original Coordinates (0,0)(2,4)(5,4)(6,6)(3,6) Coordinates after 90 o turn (0,0)(-4,2)(-4,5)(-6,6)(-6,3) Next turn of 90 o (move from Quadrant II to Quadrant III) (0,0)(-2,-4)(-5,-4)(-6,-6)(-3,-6) Next turn of 90 o (move from Quadrant III to Quadrant IV (0,0)(4,-2)(4,-5)(6,-6)(6,-3) Next turn of 90 o (move from Quadrant IV back to Quadrant I (0,0)(2,4)(5,4)(6,6)(3,6) Rule for rotating counterclockwise 90 o (x,y)  (-y,x) Yes the rule would hold for any of the quadrants

9 Problem 3.3 A Rotating 90 o counterclockwise answer for A2 PointABCDE Original Coordinates (0,0)(2,4)(5,4)(6,6)(3,6) Coordinates after 90 o turn (0,0)(-4,2)(-4,5)(-6,6)(-6,3) Next turn of 90 o (move from Quadrant II to Quadrant III) (0,0)(-2,-4)(-5,-4)(-6,-6)(-3,-6) Next turn of 90 o (move from Quadrant III to Quadrant IV (0,0)(4,-2)(4,-5)(6,-6)(6,-3) Next turn of 90 o (move from Quadrant IV back to Quadrant I (0,0)(2,4)(5,4)(6,6)(3,6) Rule for rotating counterclockwise 90 o (x,y)  (-y,x) Yes the point (0,0) the center of rotation remains unchanged. Yes, the flag and its image(s) make a symmetric design.

10 Problem 3.3 B 1 Rotating 120 o PointABCDE Original Coordinates (0,0)(2,4)(5,4)(6,6)(3,6) Next turn of 90 o (move from Quadrant II to Quadrant III) = 180 o rotation (0,0)(-2,-4)(-5,-4)(-6,-6)(-3,-6) Rule for 180 o rotation is (x,y)  (-x,-y)

11 Problem 3.3 B2 Rotating 120 o PointABCDE Original Coordinates (0,0)(2,4)(5,4)(6,6)(3,6) Coordinates after 90 o turn ( in Quadrant II) (0,0)(-4,2)(-4,5)(-6,6)(-6,3) Next turn of 90 o (move from Quadrant II to Quadrant III) (0,0)(-2,-4)(-5,-4)(-6,-6)(-3,-6) Next turn of 90 o (move from Quadrant III to Quadrant IV = 180 o rotation from Quadrant II (0,0)(4,-2)(4,-5)(6,-6)(6,-3) Next turn of 90 o (move from Quadrant IV back to Quadrant I (0,0)(2,4)(5,4)(6,6)(3,6) Rule for 180 o rotation is (x,y)  (-x,-y)

12 Rate Your Learning I will be able to identify the coordinate rules that describe rotation for turns of 90 o and 180 o.

13 Homework for Problem 3.3 ACE p. 61 #6 and #7


Download ppt "Linear Algebra Problem 3.3 Friday, September 5. Problem 3.2 answers."

Similar presentations


Ads by Google