a Start by drawing a line through A that is perpendicular to the line of reflection, y=x.
B
b Start by drawing a line through A' that is perpendicular to the line of reflection, y=- x.
C
c Start by drawing a line through A'' that is perpendicular to the line of reflection, y= x.
D
d Start by drawing a line through A''' that is perpendicular to the line of reflection, y= - x.
E
e Use the answer in Part D.
A
a A'(3,1)
B
b A''(- 1,- 3)
C
c A'''(- 3,- 1)
D
d A''''(1,3)
E
e They are the same point.
Practice makes perfect
a We will first draw a line through A that is perpendicular to the line of reflection, y=x. Notice that the lines y=x, and y=- x are perpendicular to each other.
m_(y=x) * m_(y=- x) = 1 * (- 1)=- 1
Hence, we need to draw a line through A that is parallel to the line y=- x. Then, we will find the horizontal and vertical changes, Δ y and Δ x, from the vertex to the intersection point of the two lines. Let's do it.
We see that Δ x =1 and Δ y=- 1. Therefore, we will go 1 unit right and 1 unit down from the intersection point T.
As a result, we get the point A'(3,1), which is the image of A.
R_(y=x)(A(1,3))=A'(3,1)
b We want to reflect the point A'(3,1) across the line y=- x.
R_(y=- x)(A')
To do it, we will first draw a line through A' that is perpendicular to the line of reflection, y=- x. Since the lines y=x, and y=- x are perpendicular to each other, we will draw a line through A' that is parallel to y=x. Then, we will find the horizontal and vertical changes, Δ y and Δ x.
We see that Δ x =- 2 and Δ y=- 2. Now, we will go 2 units left and 2 units down from the intersection point N.
As a result, we get the point A''(- 1,- 3), which is the image of A'.
R_(y=- x)(A'(3,1)) =A''(- 1,- 3)
c Next, we will reflect the point A''(- 1,- 3) across the line y=x. R_(y=x)(A'')
To do it, we will follow the same steps followed in Part A. Let's do it.
We see that Δ x =- 1 and Δ y=1. Now, we will go 1 unit left and 1 unit up from the intersection point S.
As a result, we get the point A'''(- 3,- 1), which is the image of A''(- 1,- 3).
R_(y=x)(A''(- 1,- 3)) =A'''(- 3,- 1)
d Finally, we will reflect the point A'''(- 3,- 1) across the line y=- x. R_(y=- x)(A''')
To do it, we will do the same things we did in Part B.
We see that Δ x =2 and Δ y=2. Now, we will go 2 units right and 2 units up from the intersection point S.
As a result, we get the point A''''(1,3), which is the image of A'''(- 3,- 1).
R_(y=- x)(A'''(- 3,- 1)) =A''''(1,3)
e We see that the points A and A'''' have the same coordinates. Hence, they are the same point.
