Translate, reflect, and rotate lines

Ontario Grade 9 Mathematics

Algebra

Ontario MTH1W – Expectation C4.3

Imagine sliding a ruler across a page, flipping it over a mirror, or spinning it around a pin. Each action moves the ruler to a new position, but it is still a straight line. In mathematics, these three actions are called transformations. This lesson focuses on what happens to a line — and to its equation — when you translate (slide), reflect (flip), or rotate (turn) it on a coordinate grid. Before we dive in, recall two things from Grade 8: (1) a linear equation in the form

y = mx + b

describes a straight line, where

m

is the slope (steepness) and

b

is the y-intercept (where the line crosses the y-axis); (2) every point on a coordinate grid has an address

(x, y)

. You will use both ideas throughout this lesson.

What you will learn

Translate a line on a coordinate grid and describe how the equation of the line changes.
Reflect a line over the x-axis or y-axis and write the equation of the image line.
Rotate a line 90° or 180° about the origin and describe the effect on the line's equation.
Identify which properties of a line (slope, y-intercept) change or stay the same after each transformation.

What Is a Transformation of a Line?

A transformation moves every point of a figure to a new location according to a rule. When the figure is a line, every point on that line moves, and the result is a new line called the image. The original line is sometimes called the pre-image.

Three transformations are covered in this lesson: a translation slides the line without turning it; a reflection flips the line over a mirror line; a rotation turns the line around a fixed point called the centre of rotation. In every case the image is still a straight line.

To describe the image line precisely, you identify two things: where two points on the image are, and then write the equation

y = mx + b

for the image. You will practise this process for each transformation.

The original line is called the pre-image; the moved line is called the image.
A line stays straight after any translation, reflection, or rotation.
The image line can be described by finding where two points moved and writing a new equation.

Translation: Sliding a Line

A translation moves every point the same distance in the same direction. Think of sliding a sticky note across your desk — every corner moves identically, so the note does not tilt or flip.

On a coordinate grid, a translation is described by how many units a point moves horizontally and how many units it moves vertically. If every point moves

a

units to the right and

b

units up, then the point

(x, y)

moves to

(x + a,\ y + b)

. Moving left uses a negative value for

a

, and moving down uses a negative value for

b

.

Here is the key insight for lines: because every point shifts by exactly the same amount, the slope of the line does not change. The line just moves to a new position. Only the y-intercept changes. For example, the line

y = 2x + 1

translated 3 units up becomes

y = 2x + 4

. The slope

m = 2

is unchanged; only the intercept increased by 3.

To find the image equation after a translation: pick any two points on the original line, apply the translation rule to each point to find their new positions, and then calculate the slope and y-intercept of the line through those two new points.

\text{Translation rule: } (x,\ y) \rightarrow (x + a,\ y + b)

Every point shifts the same distance and direction — the line slides without tilting.
Slope stays the same after a translation; the line's angle does not change.
A vertical translation of $k$ units changes the y-intercept by $k$ : the image of $y = mx + b$ is $y = mx + (b + k)$ .
A horizontal translation shifts the graph left or right, which also changes the y-intercept but not the slope.

Reflection: Flipping a Line

A reflection flips every point of the line over a mirror line. The image point is the same distance from the mirror as the original point, but on the opposite side. This is exactly like looking at a line in a mirror — the image appears behind the glass.

This lesson focuses on two common mirror lines: the x-axis and the y-axis. When you reflect over the x-axis, the horizontal position of a point stays fixed but the vertical position reverses sign. So the point

(x, y)

maps to

(x, -y)

. When you reflect over the y-axis, the vertical position stays fixed but the horizontal position reverses sign. So the point

(x, y)

maps to

(-x, y)

.

What happens to the equation? Reflecting

y = mx + b

over the x-axis gives

y = -mx - b

. Both the slope and the y-intercept change sign. Reflecting

y = mx + b

over the y-axis gives

y = -mx + b

. The slope changes sign but the y-intercept stays the same. You can verify these results by applying the reflection rule to two points on the original line and writing the equation of the image line.

A useful check: after a reflection, the original line and its image are symmetric about the mirror line. If you fold the grid along the mirror line, the two lines should match up exactly.

\text{Reflection over x-axis: } (x,\ y) \rightarrow (x,\ {-y}); \quad \text{over y-axis: } (x,\ y) \rightarrow ({-x},\ y)

Reflection over the x-axis: $(x, y) \rightarrow (x, -y)$ ; the slope and y-intercept both change sign.
Reflection over the y-axis: $(x, y) \rightarrow (-x, y)$ ; only the slope changes sign.
The image line is the same distance from the mirror line as the original, on the opposite side.
After reflection, the original and image lines are mirror images of each other.

Rotation: Turning a Line

A rotation turns every point of the line around a fixed point called the centre of rotation. In this lesson the centre is the origin

(0, 0)

. The amount of turning is called the angle of rotation, measured in degrees. Rotations are described as clockwise (the direction clock hands move) or counterclockwise (the opposite direction).

For a 180° rotation about the origin, the direction does not matter because a half-turn clockwise and a half-turn counterclockwise land in the same place. The rule is:

(x, y) \rightarrow (-x, -y)

. Both coordinates change sign. Applying this to two points on the line and writing the new equation shows that the slope stays the same but the y-intercept changes sign. The image line is parallel to the original (same slope) but on the opposite side of the origin.

For a 90° counterclockwise rotation about the origin, the rule is:

(x, y) \rightarrow (-y, x)

. This is a bigger change — the x and y coordinates swap roles (with a sign change). As a result, the slope of the image line is the negative reciprocal of the original slope. For example, a line with slope

2

becomes a line with slope

-\frac{1}{2}

after a 90° rotation. The two lines are perpendicular to each other.

To find the image equation after any rotation: apply the rotation rule to two convenient points on the original line, find the new coordinates, then calculate slope and y-intercept as usual.

\text{90° CCW: } (x,\ y) \rightarrow ({-y},\ x); \quad \text{180°: } (x,\ y) \rightarrow ({-x},\ {-y})

The centre of rotation in this lesson is always the origin $(0, 0)$ .
180° rotation rule: $(x, y) \rightarrow (-x, -y)$ ; slope stays the same, y-intercept changes sign.
90° counterclockwise rotation rule: $(x, y) \rightarrow (-y, x)$ ; the slope of the image is the negative reciprocal of the original slope.
Always apply the rotation rule to at least two points, then write the equation of the image line through those points.

Comparing the Three Transformations

Now that you have seen all three transformations, it helps to compare how each one affects the slope and y-intercept of a line. A translation never changes the slope — the image is always parallel to the original. A reflection changes the sign of the slope (and sometimes the y-intercept too). A 180° rotation keeps the slope the same but changes the sign of the y-intercept. A 90° rotation changes both the slope and the y-intercept.

One property that never changes under any of these transformations is that the image is still a straight line. Transformations preserve straightness. The length of any segment of the line is also preserved — but since lines extend forever in both directions, we usually focus on slope and intercept when describing image lines.

When solving problems, always state your transformation rule first, apply it to two or more points, and then build the equation. Skipping the point-mapping step is the most common source of errors.

Translation: slope unchanged, y-intercept shifts by the vertical component of the translation.
Reflection over x-axis: slope and y-intercept both change sign.
Reflection over y-axis: slope changes sign, y-intercept stays the same.
180° rotation about origin: slope unchanged, y-intercept changes sign.
90° CCW rotation about origin: slope becomes the negative reciprocal of the original slope.

Effect of Each Transformation on Slope and Y-Intercept

Transformation	Rule for point $(x, y)$	Slope of image	Y-intercept of image
Translation $k$ units up	$(x,\ y) \rightarrow (x,\ y+k)$	Same as original ( $m$ )	$b + k$
Translation $a$ units right	$(x,\ y) \rightarrow (x+a,\ y)$	Same as original ( $m$ )	$b - ma$
Reflection over x-axis	$(x,\ y) \rightarrow (x,\ {-y})$	$-m$	$-b$
Reflection over y-axis	$(x,\ y) \rightarrow ({-x},\ y)$	$-m$	$b$ (unchanged)
Rotation 180° about origin	$(x,\ y) \rightarrow ({-x},\ {-y})$	$m$ (unchanged)	$-b$
Rotation 90° CCW about origin	$(x,\ y) \rightarrow ({-y},\ x)$	$-\frac{1}{m}$ (if $m \neq 0$ )	Changes

Worked example

Applying Two Transformations to a Line

The line

\ell

has equation

y = 3x - 2

. (a) Translate

\ell

by 4 units up and 1 unit to the left to get line

\ell_1

. Write the equation of

\ell_1

. (b) Reflect

\ell

over the y-axis to get line

\ell_2

. Write the equation of

\ell_2

.

Part (a) – Step 1: Choose two points on $\ell$
Pick two easy points. When $x = 0$ : $y = 3(0) - 2 = -2$ , so one point is $(0, -2)$ . When $x = 1$ : $y = 3(1) - 2 = 1$ , so another point is $(1, 1)$ .
$P_1 = (0,\ {-2}), \quad P_2 = (1,\ 1)$
Part (a) – Step 2: Apply the translation rule
Moving 4 units up means $b = 4$ ; moving 1 unit to the left means $a = -1$ . The rule is $(x, y) \rightarrow (x + a,\ y + b) = (x - 1,\ y + 4)$ . Apply it to both points.
$P_1' = (0 - 1,\ {-2} + 4) = ({-1},\ 2), \quad P_2' = (1 - 1,\ 1 + 4) = (0,\ 5)$
Part (a) – Step 3: Find the slope of $\ell_1$
Use the two image points. Slope $= \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}$ . This must equal 3, since translation preserves slope — and it does.
$m = \frac{5 - 2}{0 - ({-1})} = \frac{3}{1} = 3$
Part (a) – Step 4: Write the equation of $\ell_1$
The image point $(0, 5)$ is the y-intercept of $\ell_1$ because its x-coordinate is 0. So $b = 5$ .
$\ell_1 : y = 3x + 5$
Part (b) – Step 1: Apply the reflection rule to the same two points
Reflection over the y-axis uses the rule $(x, y) \rightarrow (-x, y)$ . Apply it to $P_1 = (0, -2)$ and $P_2 = (1, 1)$ .
$P_1'' = (0,\ {-2}), \quad P_2'' = ({-1},\ 1)$
Part (b) – Step 2: Find the slope of $\ell_2$
Calculate slope using the two image points. The sign of the slope flips, as expected for a reflection over the y-axis.
$m = \frac{1 - ({-2})}{{-1} - 0} = \frac{3}{{-1}} = {-3}$
Part (b) – Step 3: Write the equation of $\ell_2$
The image point $P_1'' = (0, -2)$ is on the y-axis, so it is the y-intercept. Thus $b = -2$ . This matches the rule: reflecting over the y-axis keeps the y-intercept the same.
$\ell_2 : y = {-3}x - 2$

Answer: (a)

\ell_1 : y = 3x + 5

— same slope as

\ell

, y-intercept shifted up by 7 in total (4 up and 1 left together). (b)

\ell_2 : y = -3x - 2

— slope changed sign, y-intercept unchanged.

Check: For (a): plug

(-1, 2)

into

y = 3x + 5

: CAD 3(-1) + 5 = -3 + 5 = 2 ✓. Plug

(0, 5)

: CAD 3(0) + 5 = 5

✓. For (b): plug

(-1, 1) into

y = -3x - 2

:

-3(-1) - 2 = 3 - 2 = 1

✓. Plug

(0, -2)

:

-3(0) - 2 = -2

✓.

Common mistakes and how to avoid them

Changing the slope when translating a line — for example, writing a different slope for the image after sliding the line up.

Correction: A translation only slides the line. It never tilts it. The slope of the image is always identical to the slope of the original line.

Forgetting to change the sign of the y-intercept when reflecting over the x-axis, and only changing the slope.

Correction: Reflection over the x-axis sends

(x, y)

to

(x, -y)

, so every y-value changes sign. Both the slope and the y-intercept change sign: the image of

y = mx + b

is

y = -mx - b

.

Applying the 90° rotation rule as

(x, y) \rightarrow (y, -x)

(clockwise) when counterclockwise was intended, or mixing up which coordinate gets the minus sign.

Correction: For 90° counterclockwise about the origin, the rule is

(x, y) \rightarrow (-y, x)

. Write the rule down before substituting any values to avoid sign errors.

Only transforming one point and then writing the equation — leading to an equation that might have the correct slope but the wrong intercept.

Correction: Always transform at least two points. Use the second point to calculate slope, and use either image point to find the y-intercept.

Confusing a horizontal translation with a change in slope when the line is not horizontal.

Correction: A horizontal translation shifts the entire line left or right. The slope does not change; only the y-intercept changes. Use the translation rule on two points to find the correct new intercept.

Lesson summary

A transformation moves every point of a line to a new position; the image is still a straight line.
A translation slides the line without tilting it — the slope is always preserved, and only the y-intercept changes.
A reflection over the x-axis changes the sign of both the slope and the y-intercept; a reflection over the y-axis changes only the sign of the slope.
A 180° rotation about the origin keeps the slope the same but changes the sign of the y-intercept.
A 90° counterclockwise rotation about the origin changes both the slope and the y-intercept; the image slope is the negative reciprocal of the original slope.
For any transformation, apply the point rule to two points on the original line, find the image points, and write the equation of the image line through those points.

Check your understanding

Question 1

The line

y = 4x - 3

is translated 5 units down. What is the equation of the image line?

$y = 4x + 2$
$y = 4x - 8$
$y = -4x - 3$
$y = 4x + 8$

Show answer and explanation

y = 4x - 8

A vertical translation does not change the slope. Moving 5 units down subtracts 5 from the y-intercept:

-3 - 5 = -8

. The image equation is

y = 4x - 8

.

Question 2

The line

y = -2x + 6

is reflected over the x-axis. What is the equation of the image line?

$y = 2x + 6$
$y = -2x - 6$
$y = 2x - 6$
$y = -2x + 6$

Show answer and explanation

y = 2x - 6

Reflection over the x-axis applies the rule

(x, y) \rightarrow (x, -y)

, which changes the sign of all y-values. Both the slope and y-intercept change sign: slope becomes

-(-2) = 2

and intercept becomes

-(6) = -6

. The image is

y = 2x - 6

.

Question 3

A line passes through

(0, 0)

and

(2, 4)

. After a 180° rotation about the origin, which point is the image of

(2, 4)

?

$(4, 2)$
$(-4, -2)$
$(-2, -4)$
$(2, -4)$

Show answer and explanation

(-2, -4)

The 180° rotation rule is

(x, y) \rightarrow (-x, -y)

. Applying it to

(2, 4)

gives

(-2, -4)

.

Question 4

The line

y = 3x + 1

is rotated 90° counterclockwise about the origin. The image line passes through the origin's image point

(0, 0)

. What is the slope of the image line?

$3$
$-3$
$\frac{1}{3}$
$-\frac{1}{3}$

Show answer and explanation

-\frac{1}{3}

A 90° counterclockwise rotation uses the rule

(x, y) \rightarrow (-y, x)

. Pick two points on

y = 3x + 1

:

(0, 1)

maps to

(-1, 0)

and

(1, 4)

maps to

(-4, 1)

. Slope of image

= \frac{1 - 0}{-4 - (-1)} = \frac{1}{-3} = -\frac{1}{3}

. This matches the pattern: the slope of the image is the negative reciprocal of the original slope

3

.

Key terms

Pre-image: The original line (or figure) before a transformation is applied.
Image: The new line (or figure) that results after a transformation is applied.
Translation: A transformation that slides every point of a figure the same distance in the same direction, without rotating or flipping it.
Reflection: A transformation that flips every point of a figure over a mirror line to a point the same distance on the other side.
Rotation: A transformation that turns every point of a figure by a given angle around a fixed point called the centre of rotation.
Centre of rotation: The fixed point around which a figure is rotated. In this lesson, the centre is always the origin $(0, 0)$ .
Slope: A number that describes the steepness and direction of a line, calculated as rise divided by run: $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
Y-intercept: The value of $y$ where a line crosses the y-axis; the value of $b$ in the equation $y = mx + b$ .

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Published by DoAssignment. This reviewed lesson follows Ontario Grade 9 Mathematics (MTH1W), expectation C4.3. It is a study resource, not an official curriculum publication.

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C4.3 · Translate, reflect, and rotate lines

Ontario Grade 9 Mathematics

Ontario MTH1W – Expectation C4.3

What you will learn

What Is a Transformation of a Line?

Translation: Sliding a Line

Reflection: Flipping a Line

Rotation: Turning a Line

Comparing the Three Transformations

Effect of Each Transformation on Slope and Y-Intercept

Applying Two Transformations to a Line

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through MTH1W

About this lesson and its review