Reflection over a Vertical Line

Equal Distances on Either Side of a Vertical Mirror

Reflection over the vertical line $x = k$ moves each point of an object to its reflected position. The line $x = k$ is the mirror line.

The horizontal distance from the original point to the mirror line is equal to the horizontal distance from the image point to the mirror line. The $y$-coordinate of the point does not change.

Rule for Reflection over a Vertical Line

If a point $P(x, y)$ is reflected over the line $x = k$, its image's coordinates, $P'(x', y')$, are determined by the rule:

The image of point $P(x, y)$ is $P'(2k - x, y)$. The $y$-coordinate remains the same, while the $x$-coordinate changes according to its distance from the line $x=k$.

Reflecting a Point over a Vertical Line

Determine the image of point $P(3,2)$ by reflection over the line $x=3$.

In this case, $x=3$, $y=2$, and $k=3$.

Using the rule $P'(2k - x, y)$:

The image of point $P(3,2)$ is $P'(3,2)$. The point lies on the mirror line, so its image is the point itself.

The next point does not lie on the mirror line. Determine the image of $A(1,4)$ after reflection across $x=3$.

Here, $x=1$, $y=4$, and $k=3$.

The image of point $A(1,4)$ is $A'(5,4)$.

Exercises

Each problem reflects over the vertical line $x=k$. The last one gives the image and asks for the original point, so the same formula is solved backwards.

1. Determine the image of point $P(3,2)$ by reflection over the line $x=5$.
2. A point $B(-2, 5)$ is reflected over the line $x=1$. Determine the coordinates of its image!
3. The image of a point $C(x,y)$ after reflection over the line $x=-2$ is $C'(0,3)$. Determine the coordinates of point $C$!

Solutions

1. Given $P(3,2)$ and the mirror line $x=5$. So $x=3, y=2, k=5$.

   $$\begin{aligned} x' &= 2k - x \\ &= 2(5) - 3 \\ &= 10 - 3 \\ &= 7 \end{aligned}$$

   $$y' = y = 2$$

   Thus, the image of point $P$ is $P'(7,2)$.

2. Given $B(-2, 5)$ and the mirror line $x=1$. So $x=-2, y=5, k=1$.

   $$\begin{aligned} x' &= 2k - x \\ &= 2(1) - (-2) \\ &= 2 + 2 \\ &= 4 \end{aligned}$$

   $$y' = y = 5$$

   Thus, the image of point $B$ is $B'(4,5)$.

3. Given the image $C'(0,3)$ and the mirror line $x=-2$. So $x'=0, y'=3, k=-2$.

   We know $x' = 2k - x$ and $y' = y$.

   From $y' = y$, then $y = 3$.

   From $x' = 2k - x$, then $0 = 2(-2) - x$.

   $$0 = -4 - x$$

   $$x = -4$$

   The coordinates of point $C$ are $C(-4,3)$.