Reflection (In x-axis, y-axis, x = a, y = a and the origin; Invariant Points)

300. Mirror Across x = a · how parallel lines dictate reflection coordinates

The line x = a is always the perpendicular bisector of segment PP'.

Reflecting in the vertical line x = a sends (x, y) → (2a − x, y): the y-coordinate is unchanged while x flips across a. The line x = a is the perpendicular bisector of the segment joining a point to its image.

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Selina ICSE: Reflection (In x-axis, y-axis, x = a, y = a and the origin; Invariant Points)

What this lesson covers

Try to break it

Drag P around. P' sits at (2a − x, y) — same y, x flipped across the vertical line x = a. The line x = a is always the perpendicular bisector of PP', sitting exactly halfway between them. Pull P onto x = a and P' coincides with P.

How you build it

Reflect a point across the line x = a.

Draw the vertical mirror line x = 2.
Place point P at coordinates (-1, 3).
Construct a perpendicular from P to the line x = 2.
Mark P' on the perpendicular such that the distances to the line are equal.

The proof, step by step

Prove that the line x = a is the perpendicular bisector of P and its image P prime.

Draw a perpendicular from P to the line x = a, meeting it at M.
Since x = a is vertical, PM is horizontal. M has coordinates (a, y).
Extend PM to P' such that PM = MP'.
The x-coordinate of M is a. Since M is the midpoint of PP', (x_P + x_P')/2 = a.
Solving for x_P' gives x_P' = 2a - x_P. The y-coordinate remains unchanged.

Worked example

If a point P(3, -2) is reflected in the line x = -1, what are the coordinates of its image P'?

The line x = -1 is vertical. The y-coordinate remains -2. The x-coordinate of P is 3. Distance to line is |3 - (-1)| = 4. P' is 4 units to the left of x = -1, so x = -1 - 4 = -5. Thus P'(-5, -2).

(-5, -2) — correct
(5, -2)
(-1, -6)
(1, -2)