Reflecting Figures on the Coordinate Plane | Grade 8 Math | LevelKite

You already know how to slide a figure with a translation. A reflection is a different move: it flips a figure across a line called the mirror line, or line of reflection. The flipped figure is called the image. If a point is named $A$ , its image is named $A'$ , read as A prime.

A reflection is defined by what it does to each point:

If a point is not on the mirror line, travel from the point straight to the line, meeting it at a right angle. Then keep going the same distance on the other side. Where you stop is the image.
If a point is on the mirror line, its distance from the line is $0$ , so the point is its own image.

So a point and its image are always the same distance from the mirror line.

In this lesson, $x$ and $y$ are the exact coordinates of a point, and each can be any real number. Every distance is counted in grid units and is never negative.

Start with one point. In the figure below, the horizontal axis is the x-axis, where every point has $y = 0$ . The vertical axis is the y-axis, where every point has $x = 0$ . The mirror line is the x-axis.

The point labeled $(3, 2)$ is $2$ units above the x-axis. A vertical path meets the horizontal x-axis at a right angle, so travel straight down: $2$ units to reach the x-axis, then $2$ more units to the point labeled $(3, -2)$ . That point is the image.

Moving straight down did not change the $x$ -coordinate, which is still $3$ . The $y$ -coordinate changed from $2$ to its opposite, $-2$ . The opposite of a number is the number that is the same distance from $0$ on the other side of the number line.

The point labeled $(-4, 0)$ is on the x-axis, so it is $0$ units from the mirror line. It is its own image.

The points

(3, 2)

and

(3, -2)

are each

2

units from the x-axis, on opposite sides. The point

(-4, 0)

is on the x-axis, so it is its own image.

Key concept

Reflecting across an axis

Let $(x, y)$ be the exact coordinates of any point, where $x$ and $y$ are real numbers. Read $\to$ as goes to.

Across the x-axis ( $y = 0$ ): the point moves straight up or down, so $x$ stays the same. Its distance from the x-axis is $|y|$ , the absolute value of $y$ . The image is that far away on the other side, so $y$ becomes its opposite, $-y$ .
Across the y-axis ( $x = 0$ ): the point moves straight left or right, so $y$ stays the same. Its distance from the y-axis, $|x|$ , is kept on the other side, so $x$ becomes its opposite, $-x$ .

Taking the opposite turns a positive number negative and a negative number positive, and it leaves $0$ as $0$ , so a point on the mirror line stays put. Each rule works only for the axis named with it.

\text{x-axis: } (x, y) \to (x, -y) \qquad \text{y-axis: } (x, y) \to (-x, y)

Worked example

Reflect a triangle across the x-axis

Triangle $ABC$ has vertices $A(1, 2)$ , $B(5, 2)$ , and $C(5, 5)$ . Reflect it across the x-axis, the line where $y = 0$ . Find the vertices of the image and check them.

The mirror line is the x-axis, so each vertex moves straight up or down to cross it. Keep every x-coordinate and replace every y-coordinate with its opposite.
$(x, y) \to (x, -y)$
Reflect each vertex. A reflection sends a straight side to a straight side, so the images of the three vertices, joined in the same order, form the image triangle. The y-coordinates 2, 2 and 5 are positive, so their opposites are negative.
$A(1, 2) \to A'(1, -2) \quad B(5, 2) \to B'(5, -2) \quad C(5, 5) \to C'(5, -5)$
Check against the definition. Each vertex and its image have the same x-coordinate, so one is straight above the other. Their y-coordinates have opposite signs, so they are on opposite sides of the x-axis. Their distances from the x-axis, the absolute values of the y-coordinates, are equal.
$|2| = |-2| = 2 \qquad |5| = |-5| = 5$
Check a horizontal side. A and B have the same y-coordinate, so AB is horizontal and its length is the absolute value of the difference of the x-coordinates. The reflection kept every x-coordinate, so A'B' has the same length.
$AB = |5 - 1| = 4 \qquad A'B' = |5 - 1| = 4$
Check a vertical side. B and C have the same x-coordinate, so BC is vertical and its length is the absolute value of the difference of the y-coordinates. Taking opposites changes that difference from 3 to -3, and both have absolute value 3.
$BC = |5 - 2| = |3| = 3 \qquad B'C' = |-5 - (-2)| = |-3| = 3$

Answer

The image is triangle $A'B'C'$ with vertices $A'(1, -2)$ , $B'(5, -2)$ , and $C'(5, -5)$ . Sides $A'B'$ and $B'C'$ are $4$ units and $3$ units long, the same as $AB$ and $BC$ .

Triangle

ABC

is above the x-axis and its image

A'B'C'

is below it. Each vertex and its image have the same

x

-coordinate and opposite

y

-coordinates.

In the figure above, triangle $ABC$ is the triangle above the x-axis. Its vertices are the points labeled $(1, 2)$ , $(5, 2)$ , and $(5, 5)$ . The image is the triangle below the x-axis, with vertices labeled $(1, -2)$ , $(5, -2)$ , and $(5, -5)$ .

Notice how far each vertex travels. $A$ goes from $y = 2$ to $y = -2$ , a move of $2 - (-2) = 4$ units. $C$ goes from $y = 5$ to $y = -5$ , a move of $5 - (-5) = 10$ units. A translation would move every vertex the same distance. A reflection moves each point twice its own distance from the mirror line, so the figure flips instead of sliding.

The picture helps you see the flip, but your evidence is the coordinates. Every value in the example is exact because it comes from the coordinates. A length read from a drawing would only be an estimate.

Worked example

Reflect across the y-axis

Triangle $PQR$ has vertices $P(0, 2)$ , $Q(-3, 2)$ , and $R(-3, -2)$ . Reflect it across the y-axis, the line where $x = 0$ . Find the vertices of the image and check them using only coordinates.

The mirror line is the y-axis, so a vertex can only move straight left or right. Keep every y-coordinate and replace every x-coordinate with its opposite.
$(x, y) \to (-x, y)$
Reflect each vertex. The opposite of -3 is 3, a positive number. The opposite of 0 is 0, so P is its own image: it lies on the mirror line. The y-coordinate -2 stays -2, because this reflection changes only x-coordinates.
$P(0, 2) \to P'(0, 2) \quad Q(-3, 2) \to Q'(3, 2) \quad R(-3, -2) \to R'(3, -2)$
Check against the definition. Each vertex and its image have the same y-coordinate, so they are straight across from each other. Q and R are left of the y-axis and their images are right of it. The distances from the y-axis, the absolute values of the x-coordinates, are equal.
$|0| = 0 \qquad |-3| = |3| = 3$
Check the horizontal side PQ, where both y-coordinates are 2. Taking opposites changes the difference of the x-coordinates from 3 to -3, and both have absolute value 3.
$PQ = |0 - (-3)| = |3| = 3 \qquad P'Q' = |0 - 3| = |-3| = 3$
Check the vertical side QR, where both x-coordinates are -3. The y-coordinates did not change, so the difference is the same.
$QR = |2 - (-2)| = |4| = 4 \qquad Q'R' = |2 - (-2)| = |4| = 4$

Answer

The image is triangle $P'Q'R'$ with vertices $P'(0, 2)$ , $Q'(3, 2)$ , and $R'(3, -2)$ . Vertex $P$ is on the y-axis, so it stays in place. Sides $P'Q'$ and $Q'R'$ are $3$ units and $4$ units long, the same as $PQ$ and $QR$ .

Worked example

Reflect across the vertical line x = 1

Triangle $JKL$ has vertices $J(4, 1)$ , $K(7, 1)$ , and $L(7, 5)$ . Reflect it across the vertical line $x = 1$ , which passes through every point whose $x$ -coordinate is exactly $1$ . Find the vertices of the image and check them. The figure after this example shows this mirror line as Line A.

This mirror line is not an axis, so the two axis rules do not apply. Use the definition instead. A horizontal path meets a vertical line at a right angle, so each vertex moves straight left or right and every y-coordinate stays the same.
Find how far each vertex is from the line. Along a horizontal path, that distance is the absolute value of the difference between the vertex's x-coordinate and 1.
$J: |4 - 1| = 3 \qquad K: |7 - 1| = 6 \qquad L: |7 - 1| = 6$
The x-coordinates 4, 7 and 7 are greater than 1, so all three vertices are to the right of the line. Each image is the same distance to the left of it. Moving left lowers the x-coordinate, so subtract each distance from 1.
$J': 1 - 3 = -2 \qquad K': 1 - 6 = -5 \qquad L': 1 - 6 = -5$
Pair each new x-coordinate with the unchanged y-coordinate.
$J(4, 1) \to J'(-2, 1) \quad K(7, 1) \to K'(-5, 1) \quad L(7, 5) \to L'(-5, 5)$
Check the distances of the images from the line. They match step 2. The y-axis rule would have sent K to (-7, 1), which is 8 units from this line instead of 6, so that rule fails for this mirror.
$|-2 - 1| = |-3| = 3 \qquad |-5 - 1| = |-6| = 6 \qquad |-7 - 1| = |-8| = 8$
Check the side lengths. JK is horizontal, with length 7 - 4 = 3 units. KL is vertical, with length 5 - 1 = 4 units. The image sides match. In the figure, the triangle to the right of Line A is JKL and the triangle to the left is its image, with every vertex labeled by its coordinates.
$J'K' = |-5 - (-2)| = |-3| = 3 \qquad K'L' = |5 - 1| = 4$

Answer

The image of triangle $JKL$ across the line $x = 1$ is triangle $J'K'L'$ with vertices $J'(-2, 1)$ , $K'(-5, 1)$ , and $L'(-5, 5)$ .

Line A is the mirror line

x = 1

. Triangle

JKL

is to its right and its image is to its left, with each vertex and its image the same distance from Line A.

The takeaway: a reflection flips a figure across a mirror line. A point that is not on the line and its image are straight across the line from each other (the path between them meets the line at a right angle), the same distance away on opposite sides. A point on the line stays where it is.

Across the x-axis, keep $x$ and take the opposite of $y$ .
Across the y-axis, keep $y$ and take the opposite of $x$ .
Across any other mirror line, such as $x = 1$ , go back to the definition and work with distances.

Check every image with exact coordinates, not with how the picture looks. Distances from the mirror line should match, and side lengths should not change.

Next, you will turn figures with rotations on the coordinate plane and learn to write the rule for a transformation.