Symmetric point with respect to a line

Statement

Find the symmetric point of $P(3;-2)$ with respect to the line $y = x + 1$.

Solution

The line can be written as $x - y + 1 = 0$, so $a=1$, $b=-1$, $c=1$. For the reflection of a point $(x_0;y_0)$ with respect to the line $ax+by+c=0$ we compute $t = \frac{a x_0 + b y_0 + c}{a^2 + b^2} = \frac{3 - (-2) + 1}{1^2 + (-1)^2} = \frac{6}{2} = 3.$ The symmetric point is $\begin{aligned} x' &= x_0 - 2a\,t = 3 - 2\cdot 1\cdot 3 = -3,\\ y' &= y_0 - 2b\,t = -2 - 2\cdot(-1)\cdot 3 = 4. \end{aligned}$ Hence $\boxed{P'(-3;4)}$. Check: the midpoint of $PP'$ is $(0;1)$, which lies on the line ($1 = 0 + 1$), and the line is perpendicular to $PP'$.

Links

Topics: Plane transformations
Concepts: Axial symmetry
Methods: Isometric transformations
Skills: Analytic geometry
Exercise type: Geometric problem