Reflection over Point

Understanding Reflection over a Point

Reflection over a point, often called a half-turn rotation ($180^\circ$), is a geometric transformation where each point on an object is mapped to a new position such that the center of reflection becomes the midpoint between the original point and its image.

Suppose the center of reflection is $P(a,b)$. If a point $Q(x,y)$ is reflected over point $P$, its image $Q'(x',y')$ will lie on the line passing through $Q$ and $P$, with $P$ as the midpoint of the segment $QQ'$.

Rule for Reflection over a Point

If a point $Q(x, y)$ is reflected over the point $P(a, b)$, its image's coordinates, $Q'(x', y')$, are determined by the formula:

Alternatively, it can be written as:

This means the x-coordinate of the image is twice the x-coordinate of the center minus the original x-coordinate, and the same applies to the y-coordinate.

Reflecting a Point over Another Point

Determine the image of a half-turn $\sigma_{(2,3)}$ for the point $(5, 4)$.

This means we are reflecting point $Q(5,4)$ over the center point $P(2,3)$.

Here, $x=5$, $y=4$, $a=2$, and $b=3$.

Using the formula:

Thus, the image of point $(5,4)$ is $(-1,2)$.

Reflecting a Line over a Point

Determine the image of a half-turn $\sigma_{(1,3)}$ for the line $l$ with the equation $2x - y + 3 = 0$.

Take an arbitrary point $(x,y)$ on line $l$. Its image, $(x',y')$, after reflection over point $P(1,3)$ is:

Substitute $x = 2 - x'$ and $y = 6 - y'$ into the equation of line $l$:

Replacing $x'$ and $y'$ back to $x$ and $y$, the equation of the image line $l'$ is:

Alternatively, it can be written as $2x - y - 1 = 0$.

Exercises

1. Determine the image of a half-turn $\sigma_{(-2,3)}$ for the point $(3, -4)$.
2. Point $K(-5, 1)$ is reflected over the origin $O(0,0)$. Determine the coordinates of its image!
3. Determine the image of a half-turn $\sigma_{(1,-3)}$ for the line $l$ with the equation $2x + 5y + 6 = 0$.

Key Answers

1. Center $P(-2,3)$, point $Q(3,-4)$. So $a=-2, b=3, x=3, y=-4$.

   $$x' = 2a - x = 2(-2) - 3 = -4 - 3 = -7$$

   $$y' = 2b - y = 2(3) - (-4) = 6 + 4 = 10$$

   Thus, the image of point $Q(3,-4)$ is $(-7,10)$.

2. Center $O(0,0)$, point $K(-5,1)$. So $a=0, b=0, x=-5, y=1$.

   $$x' = 2(0) - (-5) = 0 + 5 = 5$$

   $$y' = 2(0) - 1 = 0 - 1 = -1$$

   Thus, the image of point $K(-5,1)$ is $K'(5,-1)$.

3. Center $P(1,-3)$. Line $2x + 5y + 6 = 0$.

   $$x' = 2(1) - x = 2 - x \implies x = 2 - x'$$

   $$y' = 2(-3) - y = -6 - y \implies y = -6 - y'$$

   Substitute into the line equation:

   $$2(2 - x') + 5(-6 - y') + 6 = 0$$

   $$4 - 2x' - 30 - 5y' + 6 = 0$$

   $$-2x' - 5y' - 20 = 0$$

   Image line equation: $-2x - 5y - 20 = 0$ or $2x + 5y + 20 = 0$.