What is (4, −1) reflected in the line x = 6?

It is (8, −1). The mirror line x = 6 is a vertical fold, so a reflection keeps the y-coordinate fixed and moves the point to the same distance on the other side. The point (4, −1) is 2 units to the left of x = 6 (because 6 − 4 = 2), so its image is 2 units to the right, at x = 6 + 2 = 8. The image is (8, −1). Reflecting in the y-axis instead would give (−4, −1), and reflecting in the x-axis would give (4, 1), but the question named x = 6 as the mirror.

How do you reflect a point in a line like x = 6 or y = 5?

Treat the named line as the fold. For a vertical line x = a, the y-coordinate stays the same and the x-coordinate moves to the same distance on the other side of a. For (3, 2) reflected in y = 5, the line is horizontal so the x-coordinate stays at 3; the point is 3 below the line (5 − 2 = 3), so the image is 3 above it at y = 5 + 3 = 8, giving (3, 8). Always measure the distance from the point to the mirror line, then step the same distance across.

Why do students reflect in an axis instead of the given line?

Because most early practice reflects in the x-axis or y-axis, where the rule collapses to a quick sign change: negate y for the x-axis, negate x for the y-axis. When a question names a different line such as x = 6, students reach for that reflex and reflect in the nearest axis instead. The general rule still applies: the mirror line is the fold, and each point lands the same distance on the opposite side of it. The axis rules are just the special case where the fold is x = 0 or y = 0.

GCSE Maths Foundation · AQA · Transformations & vectors

Reflecting in the given line: (4, −1) reflected in x = 6 is (8, −1)

Asked to reflect $(4,\ -1)$ in the line $x = 6$ , students reflect in an axis out of habit and write (4, 1) (the x-axis) or (−4, −1) (the y-axis). But the question names $x = 6$ as the mirror, and that line is the fold. The point is 2 to the left of it, so the image is 2 to the right: (8, −1).

The thirty-second fix: the named line is the fold, so each point lands the same distance on the other side of it. For the vertical line $x = 6$ , keep the y-coordinate and move x across: $6 + (6 - 4) = 8$ , giving $(8,\ -1)$ .

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How to spot it in your own work

You reflected in the x-axis when the question named a different line, e.g. turning $(4,\ -1)$ into $(4,\ 1)$ instead of $(8,\ -1)$ .
You reflected in the y-axis instead, e.g. writing $(-4,\ -1)$ when the mirror was $x = 6$ .
You stopped at the point on the line itself, e.g. answering $(6,\ -1)$ — the foot of the fold, not the reflected image.
You changed the wrong coordinate — for a vertical mirror $x = a$ the y-coordinate must stay fixed, and only x moves.

An exam question that triggers it

Here is a canonical AQA Foundation trigger (reflect in a stated line):

The point C has coordinates

$(4,\ -1).$

Reflect C in the line $x = 6$ . Write down the coordinates of the image.

The misconception is to ignore the named line and reflect in an axis, or to land on the line itself. The mirror is $x = 6$ , a vertical line, so the y-coordinate is unchanged and the x-coordinate moves to the same distance on the far side.

C is 2 units left of the line ( $6 - 4 = 2$ ), so its image is 2 units right, at $x = 8$ . The image is $(8,\ -1)$ .

Why students fall for this

Early reflection practice almost always uses an axis as the mirror, where the work collapses to a sign change: reflect in the x-axis by negating y, reflect in the y-axis by negating x. That shortcut becomes a reflex. So when a question names a line like $x = 6$ , students reach for the nearest axis out of habit and answer $(4,\ 1)$ or $(-4,\ -1)$ — a reflection in the wrong line entirely.

Folding fixes it. The mirror line is the crease: every point and its image sit the same distance from the line, on opposite sides, along the perpendicular. For the vertical line $x = 6$ , that perpendicular is horizontal, so the y-coordinate never moves. The point $(4,\ -1)$ is $6 - 4 = 2$ to the left, so its image is 2 to the right at $x = 8$ , giving $(8,\ -1)$ .

A second slip is to stop on the line itself, answering $(6,\ -1)$ . That is the foot of the perpendicular, halfway across — the image must travel the full distance to the other side. The diagram below shows the fold and the completed reflection.

The fix: The named line is the fold, so step the same distance across it

Read the mirror line, not an axis. If the question says reflect in $x = 6$ , the fold is that vertical line, not the x-axis or y-axis. The axis rules are only the special case where the fold is $x = 0$ or $y = 0$ .

Keep the coordinate the line fixes. A vertical mirror $x = a$ leaves the y-coordinate unchanged; a horizontal mirror $y = b$ leaves the x-coordinate unchanged.

Measure the distance, then double it across. For $(4,\ -1)$ and $x = 6$ , the gap is $6 - 4 = 2$ , so the image is at $6 + 2 = 8$ : the point $(8,\ -1)$ .

Don't stop on the line. Landing at $(6,\ -1)$ is only halfway. The image must reach the same distance on the far side of the fold.

Worked example

Reflect $(4,\ -1)$ in the line $x = 6$ . The trap is to reflect in an axis; the fix is to fold across the line you were given.

Identify the mirror. The line is $x = 6$ , a vertical line, so the fold runs top to bottom and the y-coordinate will not change.
Measure the distance to the line. The x-coordinate is 4 and the line is at $x = 6$ .
$6 - 4 = 2$
Step the same distance across. Move 2 units to the other side of the line.
$6 + 2 = 8 \quad\Longrightarrow\quad (8,\ -1)$
Vary the axis to check the method. Reflect $(3,\ 2)$ in the horizontal line $y = 5$ : now x stays at 3, the gap is $5 - 2 = 3$ , so the image is $(3,\ 8)$ . Same rule, different fold.

So $(4,\ -1)$ reflected in $x = 6$ is $(8,\ -1)$ — not the $(4,\ 1)$ of an x-axis reflection, the $(-4,\ -1)$ of a y-axis reflection, or the $(6,\ -1)$ that stops on the line.

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Common questions

What is $(4,\ -1)$ reflected in the line $x = 6$ ?: It is $(8,\ -1)$ . The mirror line $x = 6$ is a vertical fold, so the reflection keeps the y-coordinate fixed and moves the point to the same distance on the other side. The point is 2 units to the left of $x = 6$ (because $6 - 4 = 2$ ), so its image is 2 units to the right, at $x = 8$ . Reflecting in the y-axis would give $(-4,\ -1)$ and reflecting in the x-axis would give $(4,\ 1)$ , but the question named $x = 6$ .
How do you reflect a point in a line like $x = 6$ or $y = 5$ ?: Treat the named line as the fold. For a vertical line $x = a$ , the y-coordinate stays the same and the x-coordinate moves to the same distance on the other side of $a$ . For $(3,\ 2)$ reflected in $y = 5$ , the line is horizontal so x stays at 3; the point is 3 below the line ( $5 - 2 = 3$ ), so the image is 3 above it at $y = 8$ , giving $(3,\ 8)$ . Always measure the distance from the point to the mirror line, then step the same distance across.
Why do students reflect in an axis instead of the given line?: Because most early practice reflects in the x-axis or y-axis, where the rule collapses to a quick sign change: negate y for the x-axis, negate x for the y-axis. When a question names a different line such as $x = 6$ , students reach for that reflex and reflect in the nearest axis instead. The general rule still applies: the mirror line is the fold, and each point lands the same distance on the opposite side of it. The axis rules are just the special case where the fold is $x = 0$ or $y = 0$ .

Related misconceptions

Enlargement and scale factorThe neighbouring transformations skill: enlarging multiplies every length by the scale factor, so a base of 6 enlarged by ⅓ becomes 2, not 8 or 18.
Column vectors and reversingThe neighbouring vectors skill: reversing a vector negates each component in place, so (−3, 7) becomes (3, −7), not (7, −3).

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