Translate Figures

Key Concepts

• Define a translation, image and preimage.
• Translate a figure in the coordinate plane.
•  Understand isometry/ congruence transformation.
•  Write a translation rule and verify congruence.
• Understand translation theorem.

Introduction

Transformation moves or changes a figure in some way to produce a new figure called an image. Another name for the original figure is preimage.

Example: 1. In the given graph, A is the image. Find the preimage.

Translation

Translation moves every point of a figure to the same distance in the same direction. More specifically, a translation maps, or moves, the points P and Q of a plane figure to the points P΄ (read “P prime”) and Q΄, so that one of the following statements is true:

• PP΄ = QQ΄ and  = , or
• PP΄ = QQ΄ and  and  are collinear.

Translate a figure in the coordinate plane:

Translation of a figure in a coordinate plane happens horizontally and vertically.

W.r.t preimage On X-axis,

Positive translation – Shifting to the right (x + h)

Negative translation – Shifting to the left (x – h)

W.r.t preimage On Y-axis,

Positive translation – Shifting vertically upward (y + k)

Negative translation – Shifting vertically downward (y – k)

If (x, y) is the preimage, then, (x ± h, y ± k) is the image.

Example 1:

Graph quadrilateral ABCD with vertices A(–1, 2), B(-1, 5), C(4, 6), and D(4, 2). Find the image of each vertex after the translation (x, y) → (x + 3, y – 1). Then graph the image using prime notation.

Solution:

First, draw ABCD. Find the translation of each vertex by adding 3 to its x-coordinate and subtracting 1 from its y-coordinate. Then graph the image.

(x, y) → (x + 3, y – 1)

A(–1, 2) → A’(2, 1)

B(–1, 5) → B’(2, 4)

C(4, 6) → C’(7, 5)

D(4, 2) → D’(7, 1)

Congruence Transformation/ Isometry

An isometry is a transformation that preserves length and angle measure. Isometry is another word for congruence transformation.

Write a translation rule and verify congruence:

Let us understand this concept with the help of an example.

By using the graph, we go to image and pre-image of the figures and verify for congruence.

Example 2:

Write a rule for the translation of Δ ABC to Δ A’B’C’. Then verify that the transformation is an isometry.

Solution:

To go from A to A’, move 4 units left and 1 unit up.

So, a rule for the translation is (x, y) → (x – 4, y + 1).

Use the SAS Congruence Postulate.

Notice that CB = C’B’ = 3, and AC = A’C’ = 2.

The slopes of  CB− = C’B’− are 0, and the slopes of CA− = C’A’−  are undefined,

So, the sides are perpendicular. Therefore, ∠ C and ∠ C’ are congruent right angles. So, Δ ABC ≅

Δ A’B’C’. The translation is an isometry.

Translation Theorem

To prove – Translation is an isometry.

Proof:

Given: P(a, b) and Q(c, d) are two points on a figure translated by (x, y) → (x + s, y + t).

The translation maps P(a, b) to P’(a + s, b + t) and Q(c, d) to Q’(c + s, d + t).

Use the Distance Formula to find PQ and P’Q’.

PQ = √(c−a)2+(d−b)2

√P′Q′ [c+s−(a+s)]2+[(d+t)−(b+t)2

= √(c+s−a−s)2+(d+t−b−t)2

  =√(c−a)2+(d−b)2

Use the Distance Formula to find PQ and P’Q’. 

PQ = √(c−a)2+(d−b)2

P′Q′= √P′Q′= [c+s−(a+s)]2+[(d+t)−(b+t)2

= √(c+s−a−s)2+(d+t−b−t)2

= √(c−a)2+(d−b)2

Therefore, PQ = P’Q’ by the Transitive Property of Equality.

Example 3:

Draw  with vertices A(2, 2), B(5, 2), and C(3, 5). Find the image of each vertex after the translation (x, y) → (x + 1, y + 2). Graph the image using prime notation.

Solution:

Given vertices of a triangle are A(2, 2), B(5, 2), and C(3, 5).

Let us plot the vertices of a triangle on the graph.

Transformation rule:

(x, y) → (x + 1, y + 2)

A(2, 2) → A’(3, 4)

B(5, 2) → B’(6, 4)

C(3, 5) → C’(4, 7)

The image of each vertex after translation are:

A’(3,4)

B’(6,4)

C’(4,7)

Now, let us plot the image points of the triangle.

Example 4:

Δ ABC is the image of ABC after a translation. Write a rule for the translation.

Solution:

On X-axis,

Positive translation – Shifting to the right (x + h)

Negative translation – Shifting to the left (x – h)

On Y-axis,

Positive translation – Shifting vertically upward (y + k)

Negative translation – Shifting vertically downward (y – k)

Here,

The figure moved to the right on the x-axis, and the figure moved upward on the y-axis.

Translation is 3 units right and 1 unit up.

Rule: (x, y) à (x + 3, y +1)

Exercise

1. The image of (x, y) → (x + 4, y – 7) is  with endpoints P’(–3, 4) and Q’(2, 1). Find the coordinates of the endpoints of the preimage.
2. ΔA’B’C’ is the image of ABC after a translation. Write a rule for the translation. Then verify that the translation is an isometry.

• Use the translation (x, y) → (x – 8, y + 4). What is the image of B(21, 5)?
• Use the translation (x, y) → (x – 8, y + 4). What is the preimage of C’(23, 10)?
• The vertices of PQR are P(–2, 3), Q(1, 2), and R(3, –1). Graph the image of the triangle using prime notation. (x, y) → (x + 9, y – 2)
• The vertices of PQR are P(–2, 3), Q(1, 2), and R(3, –1). Graph the image of the triangle using prime notation. (x, y) → (x – 2, y – 5)
• Translate Q(0, –8) using (x, y) → (x – 3, y + 2).
• The vertices of ABC are A(2, 2), B(4, 2), and C(3, 4). a. Graph the image of ABC after the transformation (x, y) → (x +1, y). Is the transformation an isometry?
• The vertices of JKLM are J(–1, 6), K(2, 5), L(2, 2), and M(–1, 1). Graph its image after the transformation described. Translate 3 units left and 1 unit down.
• The vertices of JKLM are J(–1, 6), K(2, 5), L(2, 2), and M(–1, 1). Graph its image after the transformation described. Translate 3 units right and 1 unit up.

Concept Map

What we have learned

• Translating a figure in the coordinate plane.
• Congruence transformation.
•  Writing a translation rule and verifying congruence.
•  Translation Theorem