Big Ideas Math Geometry, 2014
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Big Ideas Math Geometry, 2014 View details
3. Rotations
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Exercise 29 Page 195

Practice makes perfect
a To rotate a line we need to know just two points on the line. To keep things simple, we can rotate the x-and y-intercept of the line. Since the slope-intercept form is y=2x-3, we know that the y-intercept is at (0,- 3). To find the x-intercept we set y equal to 0 and solve for x.
y=2x-3
0=2x-3
Solve for x
3=2x
2x=3
x=1.5
The x-intercept is at (1.5,0). To rotate a point by 90^(∘), 180^(∘), 270^(∘) and 360^(∘) about the origin, there are a number of rules that govern the new location of such a point. 90^(∘) Rotation:& (a,b)→ (- b,a) 180^(∘) Rotation:& (a,b)→ (- a,- b) 270^(∘) Rotation:& (a,b)→ (b,- a)

Let's apply these rules on the x- and y-intercept of the line.

Rotation Formula x-intercept Transformation y-intercept Transformation
90^(∘) (a,b)→ (- b,a) (1.5,0) (0,1.5) (0,- 3) (3,0)
180^(∘) (a,b)→ (- a,- b) (1.5,0) (- 1.5,0) (0,- 3) (0,3)
270^(∘) (a,b)→ (b,- a) (1.5,0) (0,- 1.5) (0,- 3) (- 3,0)

Rotating something 360^(∘) about the origin will simply bring whatever is being rotated back to its original place. In other words, when we rotate the line 360^(∘), the x- and y-intercept will stay the same. Now we can graph each rotation.

From the graphs, we already know the y-intercept of each line that's been rotated. Additionally, we know that a 360^(∘) rotation maps the line onto itself so it's slope is m=2. So far we can write the different lines as. 90^(∘) Rotation:& y=mx+1.5 180^(∘) Rotation:& y=mx+3 270^(∘) Rotation:& y=mx-1.5 360^(∘) Rotation:& y= 2x-3 To complete the equations, we use the Slope Formula.

Points y_2-y_1/x_2-x_1 m
90^(∘) ( 3,0) & ( 0,1.5) 0- 1.5/3- 0 - 1/2
180^(∘) ( 0,3) & ( - 1.5,0) 3- 0/0-( - 1.5) 2
270^(∘) ( 0,- 1.5) & ( - 3,0) - 1.5- 0/0-( - 3) - 1/2

Now we can write the equation of the line of each image. 90^(∘) Rotation:& y=- 1/2x+1.5 [0.8em] 180^(∘) Rotation:& y= 2x+3 [0.6em] 270^(∘) Rotation:& y=- 1/2x-1.5 [0.8em] 360^(∘) Rotation:& y= 2x-3 Since 2 and - 12 are negative reciprocals, we know that a rotation of 90^(∘) or 270^(∘) results in an image and preimage that are perpendicular. Additionally, when we do a 180^(∘) rotation, the preimage and image will be parallel. Finally, a 360^(∘) rotation results in an image and preimage that overlap.

b Yes the relationships are always true.

Rotation of 90^(∘) and 270^(∘)

A rotation of 90^(∘) or 270^(∘) will result in two perpendicular lines. We can prove this generally by using the Slope Formula. When rotating a line by 90^(∘), the points on the line will transform according to the following rule.
(a,b)→ (- b,a) Let's imagine an arbitrary line that intersects the y- and x-axis at A( 0,a) and B( b,0) This line will have a slope m_1. m_1=a- 0/0- b = - a/b If we rotate this line 90^(∘), applying the formula to our x- and y-intercepts, we can find the image x- and y-intercepts of the new lines.
Point (a,b) (- b,a)
A (0,a) (- a,0)
B (b,0) (0,b)
Let's calculate the slope m_2 between ( - a,0) and ( 0,b). m_2=0- b/- a- 0 = b/a If the lines are perpendicular, the product of the two lines slopes should equal - 1.
m_1* m_2 = - 1
- a/b* b/a ? = - 1
- ab/ab ? = - 1
- 1 = - 1
The lines are perpendicular. We could prove that a 270^(∘) rotation also results in a perpendicular line in a similar way.

Rotation of 180^(∘)

When rotating a line by 180^(∘), the points on the line will transform according to the following rule. (a,b)→ (- a,- b) Again, let's imagine an arbitrary line that intersects the y- and x-axes at A( 0,a) and B( b,0). We can find the slope of this line. m_1=a- 0/0- b = - a/b If we rotate this line 180^(∘) and apply the formula to our x- and y-intercept, we can find the image x- and y-intercept.

Point (a,b) (- a,- b)
A (0,a) (0,- a)
B (b,0) (- b,0)

Let's calculate the slope between ( 0,- a) and ( - b,0). m_2=- a- 0/0-( - b) = - a/b As you can see, the two lines have the same slope and are parallel.