Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Lesson 12.4 Answer Key Algebraic Representations of Transformations.

## Texas Go Math Grade 8 Lesson 12.4 Answer Key Algebraic Representations of Transformations

**Your Turn**

Question 1.

A rectangle has vertices at (0, -2), (0, 3), (3, -2), and (3, 3). What are the coordinates of the vertices of the image after the translation (x, y) → (x – 6, y – 3)? Describe the translation.

Answer:

Coordinates of the vertices of the image after the translation are:

A'(-6, -5);

B'(-6, 0);

C'(-3, -5);

D'(-3, 0).

The translation is 6 units to the left and 3 units down.

Question 2.

Triangle ABC has vertices A(-2, 6), 8(0, 5), and C(3, -1). Find the vertices of triangle A’B’C’ after a reflection across the x-axis.

Answer:

The vertices of triangle A’B’C’ after a reflection across the x-axis are:

A'(-2, -6);

B'(0, -5);

C'(3, 1).

**Reflect**

Question 3.

Communicate Mathematical Ideas How would you find the vertices of an image if a figure were rotated 270° clockwise? Explain.

Answer:

If a figure is rotated 270° clockwise, it’s the same like the figure is rotated 90° counterclockwise. Multiply each y- coordinate by 1, then switch the x- and y-coordinates: (x, y) will be (-y, x).

**Your Turn**

Question 4.

A triangle has vertices at j(-2, -4), K(1, 5), and L(2, 2). What are the coordinates of the vertices of the image after the triangle is rotated 90° counterclockwise?

Answer:

Step 1: Apply the rule to find the vertices of the image. Multiply each y-coordinate by 1, then switch x- and y coordinates. (x, y) will be (-y, x).

For J(-2, -4) the image is J'(4, -2);

For K(1, 5) the image is K'(-5, 1);

For L(2, 2) the image is L'(-2, 2).

Step 2: Graph the triangle and its image.

**Texas Go Math Grade 8 Lesson 12.4 Guided Practice Answer Key**

Question 1.

Triangle XYZ has vertices X(-3, -2), Y(-1, 0), and Z(1, -6). Find the vertices of triangle X’Y’Z’ after a translation of 6 units to the right. Then graph the triangle and its image. (Example 1)

Answer:

Step 1: Apply the rule to find the vertices of the image. Add 6 to the x-coordinate (x, y) will be (x + 6, y).

For X(-3, -2) the image is X'(3, -2);

For Y(-1, 0) the image is Y'(5, 0);

For Z(1, -6) the image is Z'(7, -6).

Step 2: Graph the triangle and its image.

Question 2.

Describe what happens to the x- and y-coordinates after a point is reflected across the x-axis. (Example 2)

Answer:

After the point is reflected across the x-axis, the x coordinate does not change and the y coordinate is multiplying by 1.

(x, y) will be (x, -y)

Question 3.

Use the rule (x, y) → (y, -x) to graph the image of the triangle at right. Then describe the transformation. (Example 3)

Answer:

Step 1 Apply the rule to find vertices of the image.

A(-4, 2) → A'(2, 4)

B(2, 3) → B'(3, -2)

C(-3, 4) → C'(4, 3)

This means that transformation is rotation 90° clockwise.

Step 2: Graph the triangle and its image

**Essential Question Check-In**

Question 4.

How do the x- and y-coordinates change when a figure is translated right a units and down b units?

Answer:

The x-coordinates increase a units, and y-coordinates decrease b units.

**Texas Go Math Grade 8 Lesson 12.4 Independent Practice Answer Key**

**Write an algebraic rule to describe each transformation. Then describe the transformation.**

Question 5.

Answer:

The point (x, y) changes to (x – 2, y – 5). This is an algebraic rule.

The transformation is a translation of 2 units left and 5 units down.

Question 6.

Answer:

The point (x, y) changes to (-x, -y). This is an algebraic rule.

The transformation is a rotation for 180°.

Question 7.

Triangle XYZ has vertices X(6, -2.3), Y(7.5, 5), and Z(8, 4). When translated, X’ has coordinates (2.8, -1.3). Write a rule to describe this transformation. Then find the coordinates of Y’ and Z’.

Answer:

X(6, -2.3) → X'(2.8, -1.3)

Find how changes coordinate x.

6 – a = 2.8

a = 6 – 2.8

a = 3.2

And how changes coordinate y.

-2.3 + b = – 1.3

b = 1.3 – (-2.3)

b = -1.2 + 2.3

b = 1

The rule is (x, y) → (x – a, y + b) where a = 3.2 and b = 1.

Apply the rule to find points Y’ and Z’.

Y(7.5, 5) → Y'(7.5 – 3.2, 5 + 1) So, Y’ has coordinates (4.3, 6).

Z(8, 4) → Z'(8 – 3.2, 4 + 1). So, Z’ has coordinates (4.8, 5).

Question 8.

Point L has coordinates (3, -5). The coordinates of point L’ after a reflection are (-3, -5). Without graphing, tell which axis point L was reflected across. Explain your answer.

Answer:

The reflection is across the y-axis, because coordinate x of point L changed sign.

Question 9.

Use the rule (x, y) → (x – 2, y – 4) to graph the image of the rectangle. Then describe the transformation.

Answer:

Apply the ruLe (x, y) → (x – 2, y – 4) to find the vertices of the image A’B’C’D’.

A(1, 1) → A'(-1, -3)

B(4, 1) → B'(2, -3)

C(4, -2) → C'(2, -6)

D(1, -2) → D'(-1, -6)

The transformation is a translation 2 units left and 4 units down.

Graph the rectangle A’B’C’D’.

Question 10.

Parallelogram ABCD has vertices A(-2, -5\(\frac{1}{2}\)), 6(-4, -5\(\frac{1}{2}\)), C(-3, -2), and D(- 1, -2). Find the vertices of parallelogram A’B’C’D’ after a translation of 2\(\frac{1}{2}\) units down.

Answer:

The transformation is a translation of 2\(\frac{1}{2}\) down.

This means that y coordinate change.

Subtract 2\(\frac{1}{2}\) from the y coordinate.

Question 11.

Alexandra drew the logo shown on half-inch graph paper. Write a rule that describes the translation Alexandra used to create the shadow on the letter A.

Answer:

The translation is 1 unit right and 0.5 unit down.

So, add 1 to the x coordinate and subtract 0.5 from the y coordinate.

The rule is:

(x, y) → (x + 1, y – 0.5)

One unit is one half inch.

So, the rule also can be add 0.5 inch (1 unit) to the x coordinate and subtract 0.25 inch (0.5 unit) from the y coordinate.

(x, y) → (x + 0.5 in, y – 0.25 in)

Question 12.

Kite KLMN has vertices at K(1, 3), L(2, 4), M(3, 3), and N(2, 0). After the kite is rotated, K’ has coordinates (-3, 1). Describe the rotation, and include a rule in your description. Then find the coordinates of L’, M’, and N’.

Answer:

K(1, 3) → K'( 3, 1)

From the point K’ we can see that the rule is:

(x, y) → (y, x)

and this is a rule for the rotation 90° counterclockwise.

Apply it to all points.

L(2, 4) → L'(-4, 2)

M(3, 3) → M'(-3, 3)

N(2, 0) → N'(0, 2)

**H.O.T. Focus on Higher Order Thinking**

Question 13.

**Make a Conjecture** Graph the triangle with vertices (-3, 4), (3, 4), and (-5, -5). Use the transformation (y, x) to graph its image.

a. Which vertex of the image has the same coordinates as a vertex of the original figure? Explain why this is true.

Answer:

The point C’ has the same coordinates as a vertex C, because the transformation is:

(x, y) → (y, x) and vertex C has coordinates where x = y. If we replace them, we will get the same point C (-5, -5)

b. What is the equation of a line through the origin and this point?

Answer:

First find coefficient k, if:

The equation of a line through two points (in this case O and C’) is:

y – y_{1} = k(x – x_{1})

y – 0 = 1 (x – 0)

y = x

c. Describe the transformation of the triangle.

Answer:

This transformation is a reflection across the line y = x, because the triangle ABC and A’B’C’ are the same distance from the line y = x, which is a line of reflection.

Question 14.

**Critical Thinking** Mitchell says the point (0, 0) does not change when reflected across the x- or y-axis or when rotated about the origin. Do you agree with Mitchell? Explain why or why not.

Answer:

Yes, applying the algebraic rules for the reflection and the rotation on the point O(0, 0), we will get the same point O.

For example reflection across the x-axis:

(x, y) → (x, -y)

(0, 0) → (0, 0)

This also applies to other transformations.

Question 15.

**Analyze Relationships** Triangle ABC with vertices A(-2, -2), B(-3, 1), and C(1, 1) is translated by (x, y) → (x – 1, y + 3). Then the image, triangle A’B’C’, is translated by (x, y) → (x + 4, y – 1), resulting in A”B”C”.

a. Find the coordinates for the vertices of triangle A”B”C”.

Answer:

First find the coordinates for the vertices B’ and C’.

Apply the rule (x, y) → (x – 1, y + 3).

A(- 2, 2) → A'(-3, 1)

B(-3, 1) → B'(-4, 4)

C(1, 1) → C'(0, 4)

Then apply the rule (x, y) → (x + 4, y – 1) on the coordinates of vertices A’, B’ and C’ to find the coordinates of vertices A”, B” and C”.

A'(- 3, 1) → A”(1, 0)

B'(-4, 4) → B”(0, 3)

C'(0, 4) → C”(4, 3)

b. Write a rule for one translation that maps triangle ABC to triangle A”B”C”.

Answer:

Subtract 1 from the coordinate x and then add 4.

Add 3 to the coordinate y, then subtract 1.

(x, y) → (x – 1 + 4, y + 3 – 1) (Simplify)

(x, y) → (x + 3, y + 2)