7.6 Reflection with respect to the line $y=x$

1.1 Angle of inclination of a line
1.2 Slope or Gradient of a line

3.1 Point-slope form
3.2 Slope-Intercept form
3.3 Two point form
3.4 Determinant form
3.5 Intercept form
3.6 Normal form
3.7 Parametric form

7.1 Reflection with respect to $X-$axis
7.2 Reflection with respect to $Y-$axis
7.3 Reflection with respect to Origin
7.4 Reflection with respect to the line $x=a$
7.5 Reflection with respect to the line $y=b$
7.6 Reflection with respect to the line $y=x$

8.1 Acute (internal) angle bisector and Obtuse (external) angle bisector

Let $P(\alpha ,\beta )$ be any point and $Q({x_1},{y_1})$ be its image about the line $y=x$ and $M$ is the mid-point of $P$ and $Q$.

$$\begin{equation} \begin{aligned} \because PQ \bot RS \\ {\text{Slope of }}PQ \times {\text{Slope of }}RS = - 1 \\ \frac{{{y_1} - \beta }}{{{x_1} - \alpha }} \times 1 = - 1 \\ {x_1} - \alpha = \beta - {y_1}...(1) \\\end{aligned} \end{equation} $$

and mid-point of $PQ$ lie on $y=x$ i.e.,

$$\begin{equation} \begin{aligned} \frac{{{y_1} + \beta }}{2} = \frac{{{x_1} + \alpha }}{2} \\ {x_1} + \alpha = \beta + {y_1}...(2) \\\end{aligned} \end{equation} $$

Solving $(1)$ and $(2)$, we get $$\begin{equation} \begin{aligned} {x_1} = \beta {\text{ and }}{y_1} = \alpha \\ Q \equiv (\beta ,\alpha ) \\\end{aligned} \end{equation} $$

which shows that in order to find the reflection with respect to the line $y=x$, interchange the $x$ and $y$ coordinates.