# 10.5: Translating and Reflecting

**At Grade**Created by: CK-12

## Learning Objectives

- Graph a
*translation*in a coordinate plane. - Recognize that a
*translation*is an*isometry.* - Find the
*reflection*of a point in a line on a coordinate plane. - Verify that a
*reflection*is an*isometry.*

## Translations

A **translation** moves *every* point a given *horizontal* distance and/or a given *vertical* distance.

- When a point is moved a certain distance horizontally and/or vertically, the move is called a ______________________________.

For example, if a **translation** moves point \begin{align*}A (3, 7)\end{align*}*right* and 4 units *up* to \begin{align*}A^\prime(5, 11)\end{align*}**translation** moves *every* point in a larger figure the *same way.*

*The symbol next to the letter \begin{align*}A^\prime\end{align*} A′ above is called the*

*prime**symbol.*

*The* *prime**symbol looks like an apostrophe like you may use to show possessive, such as, “that is my brother’s book.”*

*(The apostrophe is before the s in brother’s)*

*In math, we use the* *prime**symbol to show that two things are related.*

*In the* *translation**above, the original point is related to the translated point, so instead of renaming the translated point, we use the* *prime**symbol to show this.*

The *original* point (or figure) is called the **preimage** and the *translated* point (or figure) is called the **image.** In the example given above, the **preimage** is point \begin{align*}A(3, 7)\end{align*}**image** is point \begin{align*}A^\prime (5, 11)\end{align*}**image** is *designated* (or *shown*) with the **prime** symbol.

- Another name for the
*original*point is the __________________________. - Another name for the
*translated*point is the __________________________. - The
*translated*point uses the _______________ symbol next to its naming letter.

**Example 1**

*The point \begin{align*}A (3, 7)\end{align*} A(3,7) in a*

*translation**becomes the point \begin{align*}A^\prime (2, 4)\end{align*}*A′(2,4) . What is the

*image**of*\begin{align*}B (-6, 1)\end{align*}

*in the same*

*translation**?*

Point \begin{align*}A\end{align*}*left* and 3 units *down* to get to \begin{align*}A^\prime\end{align*}*left* and 3 units *down*.

We *subtract* 1 from the \begin{align*}x-\end{align*}

\begin{align*}B^\prime &= ( -6 - 1, 1 - 3) = (-7, -2)\end{align*}

\begin{align*}B^\prime (-7, -2)\end{align*}**image** of \begin{align*}B (-6, 1)\end{align*}

Using the **Distance Formula,** you can notice the following:

\begin{align*}AB & = \sqrt{(-6-3)^2 + (1-7)^2} = \sqrt{(-9)^2 + (-6)^2} = \sqrt{117}\\ A^\prime B^\prime & = \sqrt{(-7-2)^2 + (-2-4)^2} = \sqrt{(-9)^2 + (-6)^2} = \sqrt{117}\end{align*}

Since the endpoints of \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{A^\prime B^\prime}\end{align*} moved the *same* distance horizontally and vertically, both segments have the *same length.*

## Translation is an Isometry

An **isometry** is a transformation in which *distance* is “preserved.” This means that the *distance* between any two points in the **preimage** (*before* the **translation**) is the *same* as the *distance* between the points in the **image** (*after* the **translation**).

- An
**isometry**is when ______________________________ is*preserved*from the**preimage**to the**image.**

As you saw in Example 1 above:

The **preimage** \begin{align*}AB =\end{align*} the **image** \begin{align*}A^\prime B^\prime\end{align*} (since they are both equal to \begin{align*}\sqrt{117}\end{align*})

Would we get the same result for any other point in this translation? The answer is yes. It is clear that for any point \begin{align*}X\end{align*}, the distance from \begin{align*}X\end{align*} to \begin{align*}X^\prime\end{align*} will be \begin{align*}\sqrt{117}\end{align*}. Every point moves \begin{align*}\sqrt{117}\end{align*} units to its **image.**

This is true in general:

**Translation Isometry Theorem**

Every **translation** in the coordinate plane is an **isometry.**

- Every translation in an \begin{align*}x-y\end{align*} coordinate plane is an ________________________.

## Reflection in a Line

A **reflection** in a line is as if the line were a *mirror:*

- When an object is
**reflected**in a line, the line is like a ______________________.

An object **reflects** in the mirror, and we see the **image** of the object.

- The
**image**is the*same*distance behind the mirror line as the object is in front of the mirror line. - The “line of sight” from the
*object*to the*mirror*is**perpendicular**to the mirror line itself. - The “line of sight” from the
*image*to the*mirror*is also**perpendicular**to the mirror line.

## Reflection of a Point in a Line

Point \begin{align*}P^\prime\end{align*} is the **reflection** of point \begin{align*}P\end{align*} in line \begin{align*}k\end{align*} if and only if line \begin{align*}k\end{align*} is the **perpendicular bisector** of \begin{align*}\overline{PP^\prime}\end{align*}.

- The mirror line is a perpendicular ____________________________ of the line that connects the
*object*to its reflected image.

Reflections in Special Lines

In a coordinate plane there are some “special” lines for which it is relatively easy to create **reflections:**

- the \begin{align*}x-\end{align*}axis
- the \begin{align*}y-\end{align*}axis
- the line \begin{align*}y = x\end{align*} (this line makes a \begin{align*}45^\circ\end{align*} angle between the \begin{align*}x-\end{align*}axis and the \begin{align*}y-\end{align*}axis)
- The _________-axis, the _________-axis, and the line _________ = _________ are “special” lines to use as
*mirrors*when finding**reflections**of figures.

We can develop simple formulas for reflections in these lines.

Let \begin{align*}P (x, y)\end{align*} be a point in the coordinate plane:

We now have the following reflections of \begin{align*}P (x, y):\end{align*}

- Reflection of \begin{align*}P\end{align*} in the \begin{align*}x-\end{align*}axis is \begin{align*}Q (x, –y)\end{align*}

[ the \begin{align*}x-\end{align*}coordinate stays the *same,* and the \begin{align*}y-\end{align*}coordinate is *opposite*]

- Reflection of \begin{align*}P\end{align*} in the \begin{align*}y-\end{align*}axis is \begin{align*}R (-x, y)\end{align*}

[ the \begin{align*}x-\end{align*}coordinate is *opposite,* and the \begin{align*}y-\end{align*}coordinate stays the *same* ]

- Reflection of \begin{align*}P\end{align*} in the line \begin{align*}y = x\end{align*} is \begin{align*}S (y, x)\end{align*}

[ switch the \begin{align*}x-\end{align*}coordinate and the \begin{align*}y-\end{align*}coordinate ]

Look at the graph above and you will be convinced of the first two **reflections** in the axes. We will prove the third **reflection** in the line \begin{align*}y = x\end{align*} on the next page.

- Reflections in the \begin{align*}x-\end{align*}axis have the same ________-coordinate, but the \begin{align*}y-\end{align*}coordinate has the _________________________ value.
- Reflections in the \begin{align*}y-\end{align*}axis have an ___________________________ \begin{align*}x-\end{align*}coordinate, and the \begin{align*}y-\end{align*}coordinate stays the ____________________.
- For reflections in the \begin{align*}y = x\end{align*} line, __________________ the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}coordinates.

**Example 2**

*Prove that the reflection of point \begin{align*}P (h, k)\end{align*} in the line \begin{align*}y = x\end{align*} is the point \begin{align*}S (k, h)\end{align*}.*

Here is an “outline” proof:

First, we know the **slope** of the line \begin{align*}y = x\end{align*} is 1 because \begin{align*}y = 1x + 0\end{align*}.

Next, we will investigate the **slope** of the line that connects our two points, \begin{align*}\overline{PS}\end{align*}. Use the slope formula and the values of the points’ coordinates given above:

**Slope** of \begin{align*}\overline{PS}\end{align*} is \begin{align*}\frac{k - h}{h - k} = \frac{-1 (h - k)}{h - k} = -1\end{align*}

Therefore, we have just shown that \begin{align*}\overline{PS}\end{align*} and \begin{align*}y = x\end{align*} are **perpendicular** because the *product* of their **slopes** is –1.

Finally, we can show that \begin{align*}y = x\end{align*} is the **perpendicular bisector** of \begin{align*}\overline{PS}\end{align*} by finding the **midpoint** of \begin{align*}\overline{PS}\end{align*} :

**Midpoint** of \begin{align*}\overline{PS}\end{align*} is \begin{align*} \left ( \frac{h + k}{2}, \frac{h + k}{2} \right )\end{align*}

We know the **midpoint** of \begin{align*}\overline{PS}\end{align*} is on the line \begin{align*}y = x\end{align*} because the \begin{align*}x-\end{align*}coordinate and the \begin{align*}y-\end{align*}coordinate of the **midpoint** are the same.

Therefore, the line \begin{align*}y = x\end{align*} is the **perpendicular bisector** of \begin{align*}\overline{PS}\end{align*}.

Conclusion: The points \begin{align*}P\end{align*} and \begin{align*}S\end{align*} are **reflections** in the line \begin{align*}y = x\end{align*}.

**Example 3**

*Point \begin{align*}P (5, 2)\end{align*} is reflected in the line \begin{align*}y = x\end{align*}. The image is \begin{align*}P^\prime\end{align*}. \begin{align*}P^\prime\end{align*} is then reflected in the \begin{align*}y-\end{align*}axis. The image is \begin{align*}P^{\prime \prime}\end{align*}. What are the coordinates of \begin{align*}P^{\prime \prime}\end{align*}?*

We find one reflection at a time:

- Reflect \begin{align*}P\end{align*} in the line \begin{align*}y = x\end{align*} to find \begin{align*}P^\prime\end{align*} :

For reflections in the line \begin{align*}y = x\end{align*} we ___________________ coordinates.

Therefore, \begin{align*}P^\prime\end{align*} is (2, 5).

- Reflect \begin{align*}P^\prime\end{align*} in the \begin{align*}y-\end{align*}axis:

For reflections in the \begin{align*}y-\end{align*}axis, the \begin{align*}x-\end{align*}coordinate is _____________________ and the \begin{align*}y-\end{align*}coordinate stays the _______________________.

Therefore, \begin{align*}P^{\prime \prime}\end{align*} is (–2, 5).

## Reflections Are Isometries

Like a **translation,** a **reflection** in a line is also an **isometry.** Distance between points is “preserved” (stays the same).

- A reflection in a line is an __________________________, which means that
*distance*is*preserved.*

We will verify the **isometry** for **reflection** in the \begin{align*}x-\end{align*}axis. The proof is very similar for **reflection** in the \begin{align*}y-\end{align*}axis.

The diagram below shows \begin{align*}\overline{PQ}\end{align*} and its reflection in the \begin{align*}x-\end{align*}axis, \begin{align*}\overline{P^\prime Q^\prime}\end{align*} :

Use the Distance Formula:

\begin{align*}PQ &= \sqrt{(m-h)^2 + (n-k)^2}\\ P^\prime Q^\prime &= \sqrt{(m-h)^2 + (-n-(-k))^2} = \sqrt{(m-h)^2 + (k-n)^2}\\ &= \sqrt{(m-h)^2 + (n-k)^2}\end{align*}

So \begin{align*}PQ = P^\prime Q^\prime\end{align*}

Conclusion: When a segment is **reflected** in the \begin{align*}x-\end{align*}axis, the image segment has the *same length* as the *original* **preimage** segment. This is the meaning of **isometry.** You can see that a similar argument would apply to **reflection** in *any* line.

**Reading Check:**

1. *True or false:* Both translations and reflections are isometries.

2. *What is the meaning of the statement in #1 above?*

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

3. *If a translation rule is \begin{align*}(x + 3, y - 1)\end{align*}, in which directions is a point moved?*

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

4. *When a point or figure is reflected in a line, that line acts as a mirror.*

a. *How does the \begin{align*}x-\end{align*}axis change a point that is reflected? What do you do to the coordinates of the point in this type of reflection?*

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

b. *How does the \begin{align*}y-\end{align*}axis change a point that is reflected? What do you do to the coordinates of the point in this type of reflection?*

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

c. *How does the line \begin{align*}y = x\end{align*} change a point that is reflected? What do you do to the coordinates of the point in this type of reflection?* \begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

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## Date Created:

Feb 23, 2012## Last Modified:

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