Transforming Lines

What happens to a line when you slide it, flip it, or turn it? Each move changes the equation in a predictable way. Once you know the patterns, you can look at an equation like $y = -2x + 5$ and picture its graph right away, and you can write the equation for a line just by describing how it was moved.

Key ideas

Lines through the origin

Every line of the form $y = ax$ passes through the origin $(0, 0)$, because $a \times 0 = 0$. The number $a$ is the slope (the rate of change):

• If $a \gt 0$, the line rises from left to right.
• If $a \lt 0$, the line falls from left to right.
• The bigger the number $a$ is without its sign, the steeper the line. So $y = 3x$ and $y = -3x$ are equally steep; one rises and one falls.
• If $a = 0$, the line is $y = 0$, which is the $x$-axis.

(This is the same as $y = mx$. Your curriculum uses $a$ for this page.)

Translating up or down

A translation slides a graph without turning or flipping it. To move $y = ax$ up $b$ units, add $b$ to every $y$-value:

$$y = ax \quad\longrightarrow\quad y = ax + b$$

If $b$ is negative, the line moves down. In mapping notation, each point moves like this: $(x, y) \to (x,\ y + b)$.

The slope doesn’t change, so the new line is parallel to the original. Only the $y$-intercept changes: it moves from $0$ to $b$.

Reflecting in an axis

A reflection flips a graph over a mirror line.

• In the $x$-axis: each point $(x, y)$ goes to $(x, -y)$. The $y$-values change sign, so $y = ax$ becomes $y = -ax$.
• In the $y$-axis: each point $(x, y)$ goes to $(-x, y)$. Replacing $x$ with $-x$ gives $y = a(-x)$, which is also $y = -ax$.

For a line through the origin, both reflections give the same new line: the slope changes sign, and a rising line becomes a falling one.

Left: translating $y = 2x$ (dashed) up $3$ and down $2$. Right: reflecting $y = 2x$ in either axis gives $y = -2x$.

Rotating about the origin

A rotation turns a graph around a fixed point. If you rotate a line through the origin about the origin, it still goes through the origin, so its equation is still $y = ax$ (unless it turns all the way to vertical, $x = 0$). Only $a$ changes.

• Turning a rising line so it gets steeper makes $a$ bigger: $y = \dfrac{1}{2}x$, then $y = x$, then $y = 3x$.
• Turning it so it gets flatter makes $a$ closer to $0$.
• Turning it past the vertical or horizontal changes the sign of $a$.

A special rotation: 90°

Rotating a point 90° counterclockwise about the origin sends $(x, y)$ to $(-y, x)$. Try it on the point $(1, 2)$ on the line $y = 2x$: it goes to $(-2, 1)$. The new line passes through $(0, 0)$ and $(-2, 1)$, so its slope is

$$\frac{1 - 0}{-2 - 0} = -\frac{1}{2}$$

So rotating $y = 2x$ by 90° gives $y = -\dfrac{1}{2}x$. In general, a 90° rotation changes the slope $a$ into $-\dfrac{1}{a}$: flip the fraction and change the sign. This is called the negative reciprocal. (Rotating 90° clockwise gives the same line.) In Grade 10 you’ll use this to recognize perpendicular lines.

Left: changing $a$ rotates the line $y = ax$ about the origin. Right: a 90° rotation turns $y = 2x$ into $y = -\tfrac{1}{2}x$.

Summary

Transformation of y = ax	New equation	What happens to the graph
translate up $b$ (down if $b \lt 0$)	$y = ax + b$	same slope, $y$-intercept moves to $b$
reflect in the $x$-axis or $y$-axis	$y = -ax$	slope changes sign
rotate about the origin	$y = (\text{new } a)x$	steeper or flatter, still through $(0, 0)$
rotate 90° about the origin	$y = -\dfrac{1}{a}x$	the new line meets the old one at a right angle

Worked examples

Example 1: A translation

Describe the transformation that takes $y = 3x$ to $y = 3x - 4$. Where does the point $(2, 6)$ end up?

Solution. The slope is still $3$, and $-4$ has been added. So the line is translated down $4$ units. The new line is parallel to the old one and crosses the $y$-axis at $(0, -4)$.

The point $(2, 6)$ moves down $4$: $(2, 6) \to (2, 2)$.

Check that $(2, 2)$ is on the new line: $3(2) - 4 = 6 - 4 = 2$. ✓

Example 2: A reflection

Reflect the line $y = \dfrac{1}{2}x$ in the $x$-axis. Then reflect it in the $y$-axis. Use the point $(4, 2)$ to check both.

Solution.

In the $x$-axis: change the sign of the slope: $y = -\dfrac{1}{2}x$. The point $(4, 2)$ goes to $(4, -2)$. Check: $-\dfrac{1}{2}(4) = -2$. ✓

In the $y$-axis: replace $x$ with $-x$: $y = \dfrac{1}{2}(-x) = -\dfrac{1}{2}x$. The point $(4, 2)$ goes to $(-4, 2)$. Check: $-\dfrac{1}{2}(-4) = 2$. ✓

Both reflections give the same line, $y = -\dfrac{1}{2}x$, just as the Key ideas said.

Example 3: Rotating by changing a

Here are four lines through the origin:

$$y = 0.5x \qquad y = -4x \qquad y = 2x \qquad y = -x$$

• (a) Which lines rise, and which fall?
• (b) List them from least steep to most steep.
• (c) Write the equation of a line through the origin that is steeper than $y = 2x$ and also rises.

Solution.

(a) $y = 0.5x$ and $y = 2x$ rise (positive $a$). $y = -4x$ and $y = -x$ fall (negative $a$).

(b) Compare the sizes of $a$ without their signs: $0.5$, $4$, $2$, $1$. From least to most steep:

$$y = 0.5x, \quad y = -x, \quad y = 2x, \quad y = -4x$$

(c) Any $a$ bigger than $2$ works, for example $y = 5x$. Turning $y = 2x$ counterclockwise about the origin (toward the $y$-axis) makes it steeper.

Example 4: A 90° rotation

Rotate the line $y = 3x$ by 90° counterclockwise about the origin. Find the new equation.

Solution. Pick a point on the line: $(1, 3)$. A 90° counterclockwise rotation sends $(x, y)$ to $(-y, x)$, so $(1, 3) \to (-3, 1)$.

The new line goes through $(0, 0)$ and $(-3, 1)$:

$$\text{slope} = \frac{1 - 0}{-3 - 0} = -\frac{1}{3}$$

The new line is $y = -\dfrac{1}{3}x$.

Check with the shortcut: the negative reciprocal of $3 = \dfrac{3}{1}$ is $-\dfrac{1}{3}$. ✓

Common mistakes

Thinking a bigger $b$ makes a line steeper. In $y = ax + b$, the $b$ only slides the line up or down. The steepness comes from $a$. The lines $y = 2x + 1$ and $y = 2x + 100$ are exactly as steep as each other.

Calling $y = -2x$ a translation of $y = 2x$. The minus sign is on the slope, not added on at the end. $y = -2x$ is a reflection (it falls instead of rising). A translation down would look like $y = 2x - 2$.

Thinking $y = 3x$ and $y = -3x$ have different steepness. They’re equally steep; one rises and one falls. Compare the sizes of the slopes without their signs.

Forgetting the sign in the negative reciprocal. Rotating $y = 2x$ by 90° gives $y = -\dfrac{1}{2}x$, not $y = \dfrac{1}{2}x$. Flip and change the sign. Check with a point: $(1, 2)$ rotates to $(-2, 1)$, and $-\dfrac{1}{2}(-2) = 1$. ✓

Moving the point the wrong way. A translation up adds to the $y$-coordinate only: $(x, y) \to (x, y + b)$. The $x$-coordinate doesn’t change.

Practice

1. (Warm-up) Describe the transformation that takes $y = 4x$ to $y = 4x + 7$.

Solution

The slope stays $4$ and $7$ is added, so the line is translated up $7$ units. Its $y$-intercept moves from $(0, 0)$ to $(0, 7)$.

2. (Warm-up) Write the equation of the line $y = 5x$ after a reflection in the $x$-axis.

Solution

The slope changes sign: $y = -5x$.

Check: $(1, 5)$ reflects to $(1, -5)$, and $-5(1) = -5$. ✓

3. (Warm-up) Which line is steeper, $y = -3x$ or $y = 2x$? Which one rises?

Solution

Compare $3$ and $2$: $y = -3x$ is steeper. $y = 2x$ rises (positive slope); $y = -3x$ falls.

4. (Core) Translate $y = -\dfrac{1}{3}x$ down $2$ units. Write the new equation, and find where the point $(3, -1)$ ends up.

Solution

$$y = -\frac{1}{3}x - 2$$

The point moves down $2$: $(3, -1) \to (3, -3)$.

Check: $-\dfrac{1}{3}(3) - 2 = -1 - 2 = -3$. ✓

5. (Core) A line through the origin passes through $(2, 7)$.

• (a) Find its equation.
• (b) Write the equation after a translation up $1$ unit.
• (c) Write the equation of the original line after a reflection in the $y$-axis.

Solution

(a) Slope $= \dfrac{7 - 0}{2 - 0} = \dfrac{7}{2} = 3.5$, so $y = 3.5x$.

(b) $y = 3.5x + 1$.

(c) Replace $x$ with $-x$: $y = 3.5(-x) = -3.5x$. Check: $(2, 7)$ reflects to $(-2, 7)$, and $-3.5(-2) = 7$. ✓

6. (Core) Rotate $y = 4x$ by 90° about the origin. Write the new equation and check it with a point.

Solution

The negative reciprocal of $4$ is $-\dfrac{1}{4}$, so the new line is $y = -\dfrac{1}{4}x$.

Check: $(1, 4)$ rotates 90° counterclockwise to $(-4, 1)$, and $-\dfrac{1}{4}(-4) = 1$. ✓

7. (Core) Start with $y = \dfrac{2}{3}x$. Reflect it in the $x$-axis, then translate the result up $6$ units.

• (a) Write the final equation.
• (b) Find the $x$-intercept of the final line.

Solution

(a) Reflecting gives $y = -\dfrac{2}{3}x$. Translating up $6$ gives

$$y = -\frac{2}{3}x + 6$$

(b) Set $y = 0$:

$$\begin{aligned} 0 &= -\frac{2}{3}x + 6 \\ \frac{2}{3}x &= 6 \\ x &= 6 \times \frac{3}{2} = 9 \end{aligned}$$

The $x$-intercept is $(9, 0)$. Check: $-\dfrac{2}{3}(9) + 6 = -6 + 6 = 0$. ✓

8. (Challenge) Translate $y = 2x$ right $3$ units. Find the new equation, and explain why the result is the same as a translation down.

Solution

Move two points right $3$: $(0, 0) \to (3, 0)$ and $(1, 2) \to (4, 2)$.

The slope is still $\dfrac{2 - 0}{4 - 3} = 2$. Use $(3, 0)$ in $y = 2x + b$: $0 = 6 + b$, so $b = -6$. The new line is $y = 2x - 6$.

That’s the same as translating $y = 2x$ down $6$. A line goes on forever, so sliding it sideways lands it in the same place as sliding it down by the right amount. Here, every $1$ unit right matches $2$ units down, because the slope is $2$.

9. (Challenge) Reflect $y = 2x + 3$ in the $x$-axis, and then (starting again from $y = 2x + 3$) in the $y$-axis. Explain why the two answers are different, even though they were the same for $y = 2x$.

Solution

In the $x$-axis: every $y$-value changes sign, so replace $y$ with $-y$: $-y = 2x + 3$, which gives $y = -2x - 3$.

Check: $(1, 5)$ reflects to $(1, -5)$, and $-2(1) - 3 = -5$. ✓

In the $y$-axis: replace $x$ with $-x$: $y = 2(-x) + 3 = -2x + 3$.

Check: $(1, 5)$ reflects to $(-1, 5)$, and $-2(-1) + 3 = 5$. ✓

Both reflections change the slope from $2$ to $-2$. But the $y$-intercept $(0, 3)$ is on the $y$-axis, so reflecting in the $y$-axis leaves it alone, while reflecting in the $x$-axis sends it to $(0, -3)$. For $y = 2x$, the $y$-intercept is the origin, which stays put in both reflections, so the answers matched.