Reflection Across the Line y = x

A reflection flips the plane over a line called the mirror. Every point and its image are equidistant from the mirror, on opposite sides — like a real mirror reflection.

The Formula for Reflection Across $y = x$

To reflect a point $(x, y)$ across the line $y = x$, simply swap the coordinates:

$(x, y) \to (y, x)$

That's it. The line $y = x$ contains all points where the two coordinates are equal, and reflecting across it interchanges the role of $x$ and $y$.

Other Common Reflection Lines

• Across the $x$-axis ($y = 0$): $(x, y) \to (x, -y)$
• Across the $y$-axis ($x = 0$): $(x, y) \to (-x, y)$
• Across the origin (point reflection): $(x, y) \to (-x, -y)$
• Across $y = x$: $(x, y) \to (y, x)$
• Across $y = -x$: $(x, y) \to (-y, -x)$

For an arbitrary line, the formula is more complex — but the idea is the same: drop a perpendicular from the point to the line, then continue an equal distance on the other side.

Step-by-Step Solution

Given: A triangle with vertices $A = (1, 0)$, $B = (4, 1)$, $C = (2, 3)$. Reflect across $y = x$.

Find: The image vertices.

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Step 1 — Apply $(x, y) \to (y, x)$ to each vertex.

$A = (1, 0) \to (0, 1) = A'$

$B = (4, 1) \to (1, 4) = B'$

$C = (2, 3) \to (3, 2) = C'$

Step 2 — Verify the reflected points are equidistant from the line $y = x$.

For point $(1, 0)$, the closest point on $y = x$ is found by perpendicular projection. The perpendicular to $y = x$ has slope $-1$, so the line through $(1, 0)$ perpendicular to $y = x$ is $y - 0 = -1(x - 1)$, or $y = -x + 1$. Intersecting with $y = x$:

$x = -x + 1 \Rightarrow 2x = 1 \Rightarrow x = y = 0.5$

So the foot of the perpendicular is $(0.5, 0.5)$, which is exactly the midpoint of $(1, 0)$ and $(0, 1)$. ✓ The original and reflected points are symmetric about this midpoint.

Step 3 — Verify the triangle's shape is preserved.

Compute side lengths of the original:

• $|AB| = \sqrt{(4-1)^{2} + (1-0)^{2}} = \sqrt{10}$
• $|BC| = \sqrt{(2-4)^{2} + (3-1)^{2}} = \sqrt{8}$
• $|CA| = \sqrt{(1-2)^{2} + (0-3)^{2}} = \sqrt{10}$

For the reflected triangle:

• $|A'B'| = \sqrt{(1-0)^{2} + (4-1)^{2}} = \sqrt{10}$
• $|B'C'| = \sqrt{(3-1)^{2} + (2-4)^{2}} = \sqrt{8}$
• $|C'A'| = \sqrt{(0-3)^{2} + (1-2)^{2}} = \sqrt{10}$

All match. ✓

Step 4 — Reflections reverse orientation.

Reflections are unique among rigid motions in that they flip the orientation of figures. If the original triangle has vertices ordered counterclockwise $(A, B, C)$, the reflected triangle has them ordered clockwise — like looking in a mirror. This is why your reflection in a mirror is "left-right reversed" but not "up-down reversed."

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Answer: Reflecting $A = (1, 0)$, $B = (4, 1)$, $C = (2, 3)$ across the line $y = x$:

$\boxed{\;A' = (0, 1),\quad B' = (1, 4),\quad C' = (3, 2)\;}$

The triangle is preserved in size and shape but flipped — the orientation reverses. Each vertex is the same distance from $y = x$ as its image, on the opposite side.

Try It

• The original triangle is shown faintly; the reflected triangle is bright.
• Watch the triangles flip across the cyan diagonal line $y = x$.
• Notice that the points exactly on the line ($(0, 0)$, $(1, 1)$, etc.) stay fixed.