Question

reflection over a vertical line (x = c)
patterns found in the coordinates

1. given triangle def at d(3,5), e(6,5) and f(3,1). reflect it over the y - axis (x = 0 line). determine the coordinates of d, e, and f.

d(_, _), e(_, _), & f(_, _)

1. given def at d(3,5), e(6,5) and f(3,1). reflect it over the vertical line (x = 1 line). determine the coordinates of d, e, and f.

d(_, _), e(_, _), & f(_, _)
how did moving the vertical line 1 positive value to the right change the image?
given def at d(3,5), e(6,5) and f(3,1). reflect it over the vertical line (x = 2). determine the coordinates of d, e, and f.
d(_, _), e(_, _), & f(_, _)
how did moving the vertical line 1 positive value to the right change the image?
lets see if we can create an general formula for the line
given d(3,5), e(6,5) and f(3,1)
reflected over x = 0 d(-3,5)
given d(3,5), e(6,5) and f(3,1)
reflected over x = 1 d(_)
given d(3,5), e(6,5) and f(3,1)
reflected over x = 2 d(_)
given d(3,5), e(6,5) and f(3,1)
reflected over x = c d(_)
r_{x = c}:(x,y)=( _, _)

Explanation:

Step1: Recall reflection rule over x = 0

When reflecting a point $(x,y)$ over the y - axis ($x = 0$), the rule is $(x,y)\to(-x,y)$. For point $D(3,5)$, $D'\to(- 3,5)$; for $E(6,5)$, $E'\to(-6,5)$; for $F(3,1)$, $F'\to(-3,1)$.

Step2: Recall reflection rule over x = 1

The distance between a point $(x,y)$ and the line $x = 1$ is $d=\vert x - 1\vert$. The new x - coordinate after reflection over $x = 1$ is $x'=1-(x - 1)=2 - x$. For $D(3,5)$, $x'=2-3=-1$, so $D'(-1,5)$; for $E(6,5)$, $x'=2 - 6=-4$, so $E'(-4,5)$; for $F(3,1)$, $x'=2 - 3=-1$, so $F'(-1,1)$.

Step3: Recall reflection rule over x = 2

The new x - coordinate of a point $(x,y)$ reflected over $x = 2$ is $x'=2-(x - 2)=4 - x$. For $D(3,5)$, $x'=4 - 3 = 1$, so $D'(1,5)$; for $E(6,5)$, $x'=4 - 6=-2$, so $E'(-2,5)$; for $F(3,1)$, $x'=4 - 3 = 1$, so $F'(1,1)$.

Step4: Find general formula for reflection over x = c

The distance between a point $(x,y)$ and the line $x = c$ is $\vert x - c\vert$. The new x - coordinate after reflection over $x = c$ is $x'=c-(x - c)=2c - x$, and the y - coordinate remains the same. So the rule $R_{x = c}:(x,y)=(2c - x,y)$.

Answer:

1. $D'(-3,5),E'(-6,5),F'(-3,1)$
2. $D'(-1,5),E'(-4,5),F'(-1,1)$
3. $D'(1,5),E'(-2,5),F'(1,1)$
4. $R_{x = c}:(x,y)=(2c - x,y)$