Reflection over Line x = k

Understanding Reflection over the Line x = k

Reflection over the vertical line $x = k$ is a geometric transformation where each point of an object is mapped to a new position. The line $x = k$ acts as a mirror.

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 the Line x = k

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:

Thus, the image of point $P(x, y)$ is $P'(2k - x, y)$. Note that the y-coordinate remains the same, while the x-coordinate changes based on its distance from the line $x=k$.

Reflecting a Point over the Line x = k

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

Thus, 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.

Now, let's try another example. Determine the image of point $A(1,4)$ if it is reflected over the line $x=3$.

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

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

Exercises

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!

Key Answers

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

   $$x' = 2k - x = 2(5) - 3 = 10 - 3 = 7$$

   $$y' = y = 2$$

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

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

   $$x' = 2k - x = 2(1) - (-2) = 2 + 2 = 4$$

   $$y' = y = 5$$

   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 $0 = 2(-2) - x$.

   $$0 = -4 - x$$

   $$x = -4$$

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