90^(∘) Rotation:& (a,b)→ (- b,a)
180^(∘) Rotation:& (a,b)→ (- a,- b)
270^(∘) Rotation:& (a,b)→ (b,- a)
B
b Rotate a line that passes through (a,0) and (0,b).
A
aGraphs:
Relationships and Equations: See solution.
B
b Yes it does.
Explanation: See solution.
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.
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.
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.
