Exercise 37 Page 560

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} \cdot m_{y = - x} = 1 \cdot (- 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.

A (1, 3) \Leftrightarrow A^{''''} (1, 3)

Subchapter links

Lesson Check p.557

Practice and Problem-Solving Exercises p.557-560

Standardized Test Prep p.560

Mixed Review p.560

Exercise 37 Page 560

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