Lesson 6 of 6Linear and Non-Linear Relations

Unit 5 · Lesson 6 of 6

Translating, Reflecting and Rotating Lines

See how translating, reflecting and rotating the line y = ax changes its equation: a vertical translation changes b, a reflection in an axis changes the sign of the slope, and a 90° rotation gives slope −1/a.

7 min read8 quick checks11 practice questionsInteractiveAsk AI about this →

Key things to remember

1Translating $y = a x$ up $k$ units gives $y = a x + k$ (down $k$ : $y = a x - k$ ); the slope does not change.
2Reflecting $y = a x$ in the $x$ -axis or in the $y$ -axis gives the same line, $y = - a x$ .
3Rotating $y = a x$ ( $a \neq = 0$ ) by 90° about the origin gives $y = - \frac{1}{a} x$ ; the two slopes multiply to $- 1$ .
4Point rules: up $k$ : $(x, y + k)$ ; $x$ -axis reflection: $(x, - y)$ ; $y$ -axis reflection: $(- x, y)$ ; 90° counterclockwise rotation: $(- y, x)$ .
5To check an answer, transform one point of the original line and substitute it into the new equation.

In a drawing app or a video game, shapes get moved up, flipped like a mirror image, and turned. Every one of those moves is a transformation, and a computer carries it out by applying a rule to the coordinates of every point. When the shape is a straight line, the result is another straight line, and its equation changes in a predictable way. In this lesson you start with the simplest lines, $y = a x$ , which pass through the origin, and find out exactly what happens to $a$ (and to the $y$ -intercept) when the line is translated up or down, reflected in an axis, or rotated a quarter turn.

The starting line: y = ax

The line $y = a x$ has slope $a$ and $y$ -intercept $0$ , so it passes through the origin. One easy point on it is $(1, a)$ : for $y = 2 x$ that is $(1, 2)$ . The plan for every transformation is the same: apply the transformation's rule to a couple of points on the line, then find the equation of the line through the new points. You already know how to find an equation from points, so nothing here needs a new formula you cannot check yourself.

Fig 1The lines y = 2x, y = x, y = ½x, y = −x and y = −2x all pass through the origin; the dashed line x = 1 meets each one at the point (1, a).

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Which point is on the line

y = 5 x

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What is the

y

-intercept of the line

y = - 7 x

Vertical translation

A translation slides every point the same distance in the same direction, without turning. Moving up $k$ units sends $(x, y)$ to $(x, y + k)$ . On $y = 2 x$ , the point $(0, 0)$ moves to $(0, 3)$ and $(- 2, - 4)$ moves to $(- 2, - 1)$ when $k = 3$ . Every $y$ -value grows by 3, so the new equation is $y = 2 x + 3$ . The line has not tilted, so the slope is still 2: the new line is parallel to the old one, and only the $y$ -intercept changed, from $0$ to $3$ . Moving down 4 instead gives $y = 2 x - 4$ .

Fig 2Left: y = 2x moved up 3 to y = 2x + 3. Right: y = 2x with its reflection in the x-axis, y = −2x, and its rotation by 90 degrees about the origin, y = −½x, tracked with the point (1, 2)

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Write the equation of

y = - 3 x

after it is translated down 2 units, in the form y = mx + b.

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When

y = 2 x

is translated up 3 units, what happens to its slope?

Reflection in the x-axis or the y-axis

A reflection flips the line over a mirror line. Reflecting in the $x$ -axis keeps $x$ and changes the sign of $y$ : $(x, y) \to (x, - y)$ . The point $(1, 2)$ on $y = 2 x$ lands on $(1, - 2)$ , and the reflected line through the origin and $(1, - 2)$ has slope $- 2$ : $y = - 2 x$ . Reflecting in the $y$ -axis keeps $y$ and changes the sign of $x$ : $(x, y) \to (- x, y)$ , so $(1, 2)$ lands on $(- 1, 2)$ , which is also on $y = - 2 x$ . For a line through the origin, both reflections give the same line, $y = - a x$ . That coincidence depends on the line passing through the origin. For $y = 2 x + 3$ , reflecting in the $x$ -axis gives $y = - 2 x - 3$ , but reflecting in the $y$ -axis gives $y = - 2 x + 3$ .

Fig 3The line y = 2x + 3 reflected two ways. In the x-axis, (1, 5) goes to (1, −5) and (0, 3) to (0, −3), giving y = −2x − 3. In the y-axis, (1, 5) goes to (−1, 5) and (0, 3) stays put, giving y = −2x + 3.

Transformation	Point rule	y = 2x becomes	y = ax becomes
Translate up k	(x, y) → (x, y + k)	$y = 2 x + k$	$y = a x + k$
Translate down k	(x, y) → (x, y − k)	$y = 2 x - k$	$y = a x - k$
Reflect in the x-axis	(x, y) → (x, −y)	$y = - 2 x$	$y = - a x$
Reflect in the y-axis	(x, y) → (−x, y)	$y = - 2 x$	$y = - a x$
Rotate 90° about the origin	(x, y) → (−y, x), counterclockwise	$y = - \frac{1}{2} x$	$y = - \frac{1}{a} x$ (for $a \neq = 0$ )

How each transformation changes the line y = ax

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A reflection in the

x

-axis sends the point

(3, 6)

to which point?

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The line

y = x + 4

is reflected in the

x

-axis. What is the new equation?

Rotation by 90° about the origin

A rotation turns every point around a fixed centre, here the origin. A quarter turn counterclockwise sends $(x, y)$ to $(- y, x)$ : the point $(1, 2)$ goes to $(- 2, 1)$ . The origin stays put, so the rotated line passes through $(0, 0)$ and $(- 2, 1)$ , and its slope is $\frac{1 - 0}{- 2 - 0} = - \frac{1}{2}$ . The new line is $y = - \frac{1}{2} x$ . In general, $(1, a)$ goes to $(- a, 1)$ , giving slope $\frac{1}{- a} = - \frac{1}{a}$ , the negative reciprocal of $a$ . Turning clockwise instead sends $(1, 2)$ to $(2, - 1)$ , which is on the same line, so for a line through the origin the direction of the quarter turn does not matter. The original and rotated lines meet at a right angle, and their slopes multiply to $- 1$ : $2 \times (- \frac{1}{2}) = - 1$ . One exception: if $a = 0$ , the line is the $x$ -axis, and a quarter turn makes it the $y$ -axis, $x = 0$ , which is vertical, so its slope is undefined.

y = a x rotate 9 0^{\circ} y = - \frac{1}{a} x (a \neq = 0), a \times (- \frac{1}{a}) = - 1

Loading interactive…

Open the transform panel, pick a value of a, then translate, reflect or rotate the line y = ax and watch its equation change.

Three moves, three effects

A vertical translation changes only the

y

-intercept (the slope stays the same). A reflection of

y = a x

in either axis changes the sign of the slope,

a \to - a

. A 90° rotation about the origin replaces the slope with its negative reciprocal,

a \to - \frac{1}{a}

Check by tracking a point

Not sure your new equation is right? Take one point on the original line, apply the point rule, and substitute the new point into your new equation. For

y = 3 x

rotated 90°, the point

(1, 3)

becomes

(- 3, 1)

, and

y = - \frac{1}{3} x

gives

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

✓.

Quick checkTry it before you read on

y = \frac{1}{3} x

is rotated 90° about the origin. What is the slope of the new line?

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Which pair of lines meet at a right angle?

VideoAn Ontario teacher translates, reflects and rotates lines and finds the equation of each new line. It runs about 19 minutes, so use the chapter list to jump to the summary of rules and the examples.Watch on YouTube

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Translate

y = 3 x

down 5 units. Write the new equation and check it with a point.

Example 2

Reflect

y = \frac{3}{4} x

in the

x

-axis, then in the

y

-axis. Show that both reflections give the same line.

Example 3

Rotate

y = 3 x

by 90° counterclockwise about the origin. Find the new equation.

Example 4

Rotate

y = - 2 x

by 90° about the origin, then translate the result up 4. Write the final equation and check it by tracking the point

(1, - 2)

Where marks get lost

Changing the slope when translating. Moving $y = 2 x$ up 3 gives $y = 2 x + 3$ , not $y = 5 x$ : a translation slides the line without tilting it.
Using $- a$ or $\frac{1}{a}$ for a 90° rotation. The rotated slope is the negative reciprocal, $- \frac{1}{a}$ : for $y = 4 x$ it is $- \frac{1}{4}$ .
Rotating a point with $(x, y) \to (y, x)$ . That swap is a reflection in the line $y = x$ ; a 90° counterclockwise rotation is $(x, y) \to (- y, x)$ .
Assuming the $x$ -axis and $y$ -axis reflections always give the same line. They agree for $y = a x$ , but for $y = 2 x + 3$ they give $y = - 2 x - 3$ and $y = - 2 x + 3$ .
Applying $- \frac{1}{a}$ to $y = 0$ . You cannot divide by 0: a quarter turn of the $x$ -axis gives the vertical line $x = 0$ .

Flashcards 7

Translation

y = ax translated up k

Reflection in the x-axis

Reflection in the y-axis

+ 3 more

Practice

Write the equation of

y = 4 x

after it is translated up 2 units, in the form y = mx + b.

What is the equation of

y = 5 x

after a reflection in the

x

-axis?

y = 4 x

is rotated 90° about the origin. What is the slope of the new line? Give a decimal or a fraction.

y = - \frac{2}{3} x

is rotated 90° about the origin. What is the slope of the new line? Give a decimal or a fraction.

Which transformation turns

y = 2 x

into

y = 2 x - 6

Match each transformation of

y = 3 x

to the resulting equation.

Translate up 1

Translate down 1

Reflect in the x-axis

Rotate 90° about the origin

The point

(2, 6)

is rotated 90° counterclockwise about the origin. Where does it land?

y = 2 x

is translated up 3 units. What is

y

on the new line when

x = 4

Put these steps in order to find the equation of y = ax after a 90° counterclockwise rotation about the origin.

1Rotate it with (x, y) → (−y, x) to get (−a, 1)
2Write the new equation y = −(1/a)x
3Find the slope from the origin to the new point: 1 ÷ (−a) = −1/a
4Pick a point on the line, such as (1, a)

The line

y = 2 x + 3

(which does not pass through the origin) is reflected in the

y

-axis. What is the new equation?

A line

y = a x

(with

a \neq = 0

) and its image after a 90° rotation about the origin have slopes whose product is always…

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