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10.5: Translating and Reflecting

Difficulty Level: 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.


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*}A(3,7) 2 units to the right and 4 units up to \begin{align*}A^\prime(5, 11)\end{align*}A(5,11), then this 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*}A(3,7) and the image is point \begin{align*}A^\prime (5, 11)\end{align*}A(5,11). The 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*}B(6,1) in the same translation?

Point \begin{align*}A\end{align*}A moved 1 unit to the left and 3 units down to get to \begin{align*}A^\prime\end{align*}A. Point \begin{align*}B\end{align*}B will also move 1 unit to the left and 3 units down.

We subtract 1 from the \begin{align*}x-\end{align*}xcoordinate and 3 from the \begin{align*}y-\end{align*}ycoordinate of point \begin{align*}B\end{align*}B:

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


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

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?





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





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?





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?





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*}




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