Q. How to reflect a point over \(y = x\).

Q: What is the rule to reflect a point over the line y = x?

Swap coordinates: the image of the point (x,y) is (y,x) . This follows because y = x is the line of symmetry exchanging the axes.

Q: How do I write the reflection as a matrix?

Use the matrix \begin{pmatrix}0 & 1 \\[2pt] 1 & 0\end{pmatrix} . Multiplying it by \begin{pmatrix}x\\[2pt] y\end{pmatrix} yields \begin{pmatrix}y\\[2pt] x\end{pmatrix} .

Q: How do I reflect a point across a general line ax + by + c = 0?

Let d = \frac{ax_0 + by_0 + c}{a^2 + b^2}. The reflected point is (x_0 - 2ad, y_0 - 2bd).

Q: How is this reflection expressed in the complex plane?

If z = x + iy , the reflection across y = x sends z to w = y + ix , which equals w = i\overline{z} .

Answer

how to reflect a point over y = x

For a point \((x,y)\), the reflection across the line \(y=x\) is \((y,x)\).

Quick explanation: the midpoint of \((x,y)\) and \((y,x)\) is \(\bigl(\frac{x+y}{2}, \frac{x+y}{2}\bigr)\), which lies on \(y=x\); and the slope of the segment joining them is -1, perpendicular to the slope 1 of \(y=x\), so they are symmetric.

Detailed Explanation

State the given point and the mirror line.

Let the point to be reflected be \(P=(a,b)\). The mirror line is \(y=x\).
Find the equation of the line through \(P\) that is perpendicular to \(y=x\).

The line \(y=x\) has slope 1. A line perpendicular to it has slope -1. The perpendicular line through \(P=(a,b)\) therefore has equation

\[y-b = -1(x-a)\]

or equivalently

\[y = -x + (a+b)\]
Find the intersection point \(Q\) of the perpendicular line with \(y=x\) (the foot of the perpendicular).

At the intersection, \(y=x\) and \(y=-x+(a+b)\). Solve the system:

\[x = -x + (a+b)\]

\[2x = a+b\]

\[x_Q = \frac{a+b}{2}\]

Since \(y_Q=x_Q\),

\[y_Q = \frac{a+b}{2}\]

Thus the foot is \(Q=\bigl(\frac{a+b}{2}, \frac{a+b}{2}\bigr)\).
Use the midpoint property to determine the reflected point \(P’\).

Reflection across a line makes the intersection point \(Q\) the midpoint of the segment joining \(P\) and its reflection \(P’=(x’,y’)\). Therefore

\[\frac{a+x’}{2} = x_Q = \frac{a+b}{2}\]

\[\frac{b+y’}{2} = y_Q = \frac{a+b}{2}\]

Solving these gives

\[x’ = a+b – a = b\]

\[y’ = a+b – b = a\]

Hence the reflection is \(P’=(b,a)\).
Summary and alternative matrix viewpoint.

Reflecting a point across the line \(y=x\) swaps the coordinates. For any point \(P=(a,b)\), the reflection across \(y=x\) is

\[P’=(b,a)\]

Equivalently, as a linear transformation on column vectors, the reflection is given by the matrix

\[\begin{pmatrix}0 & 1\\[4pt]1 & 0\end{pmatrix}\]

so

\[\begin{pmatrix}0 & 1\\[4pt]1 & 0\end{pmatrix}\begin{pmatrix}a\\[4pt]b\end{pmatrix}= \begin{pmatrix}b\\[4pt]a\end{pmatrix}\]
Quick numeric example.

Reflect \(P=(2,5)\) across \(y=x\). Swap coordinates to get \(P’=(5,2)\).

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FAQs

What is the rule to reflect a point over the line y = x?

Swap coordinates: the image of the point (x,y) is \( (y,x) \). This follows because y = x is the line of symmetry exchanging the axes.

How do I write the reflection as a matrix?

Use the matrix \( \begin{pmatrix}0 & 1 \\[2pt] 1 & 0\end{pmatrix} \). Multiplying it by \( \begin{pmatrix}x\\[2pt] y\end{pmatrix} \) yields \( \begin{pmatrix}y\\[2pt] x\end{pmatrix} \).

How do I reflect the graph of an equation across y = x?

Swap x and y everywhere in the equation. For example, y = 2x + 1 becomes x = 2y + 1, which can be solved for y if desired.

How do I reflect a point across the line y = x + c (parallel shifted)?

Translate the line down by c, reflect, then translate back. The direct formula: the image of \( (x,y) \) is \( (\,y - c,\; x + c\,) \).

How do I reflect a point across a general line ax + by + c = 0?

Let \(d = \frac{ax_0 + by_0 + c}{a^2 + b^2}\). The reflected point is \((x_0 - 2ad, y_0 - 2bd)\).

How do I reflect a polygon or shape over y = x?

Does reflection across y = x preserve distances and orientation?

It is an isometry, so distances and angles are preserved. Orientation is reversed (the transformation has determinant -1).

How is this reflection expressed in the complex plane?

If \( z = x + iy \), the reflection across y = x sends \( z \) to \( w = y + ix \), which equals \( w = i\overline{z} \).

Mirror points across the line y = x.
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