Jtdylktuy

I know how to reflect a coordinate over the $y$ and $x$ axis, but is there a rule I could use to help me find the reflected point over $x = -1$?

This is what I know already:

Over the $x$-axis: $(x, y) to (x, –y) $

Over the $y$-axis: $(x, y) to (–x, y)$

Over the line $y = x$: $(x, y) to (y, x)$

Through the origin: $(x, y) to (–x, –y) $

What I don't know is how to solve when reflecting over something like $x = -1$. Is there a rule that would help make solving this easier?

Reflecting a point over $x=-1$ just means you have to keep the ordinate and let $x'-(-1)=-1-x implies x'=-2-x$

Rather than think about transformation rules symbolically, and trying to generalize them, try thinking about them visually. The line $x = -1$ is a vertical line one unit to the left of the $y$-axis. (You should follow along and draw things out on a sheet of graph paper or on your computer, in order to make them clear.) Therefore, if you have a point at $(3, -5)$, it is three units to the right of the $y$-axis, but four units to the right of your axis of symmetry.

When you reflect this point, you should end up at the same "height" ($y$-coordinate) of $-5$, but this time four units to the left of your axis of symmetry. Four units to the left of $x = -1$ is $x = -1-4 = -5$, so the point $(3, -5)$ reflects to $(-5, -5)$. We might write

$$
(3, -5) to (-5, -5)
$$

Similar reasoning shows that, for example,

$$
(-2, 4) to (0, 4)
$$
$$
(0, 0) to (-2, 0)
$$
$$
(13, 2) to (-15, 2)
$$

So in each case, the $y$-coordinate stays the same, but $3$ becomes $-5$, $-2$ becomes $0$, $0$ becomes $-2$, and $13$ becomes $-15$. If you're a perceptive sort, you might notice that the sum of each of these pairs of $x$-coordinates is $-2$, and therefore arrive at the transformation rule $x' = -2-x$, but if not, you can still reconstruct what's happening.

The point $x$ is how far to the right of your axis of symmetry? The axis of symmetry has an $x$-coordinate of $-1$, so your distance to the right is $x-(-1)$, or $x+1$. (This distance is negative if you are actually to the left of the axis.) Then we need to move to the left by that amount. We do that by subtracting $x+1$ from $-1$, to get $x' = -1-(x+1) = -1-x-1 = -2-x$.

And of course, $y' = y$. So, finally,

$$
(x, y) to (x', y') = (-2-x, y)
$$

To reflect over a vertical line, such as $x=a$, first translate so the line is shifted to the y-axis, then reflect over it, then translate back so the line is shifted to its original position.

$$(x,y)mapsto (x-a,y) mapsto (a-x,y) mapsto (2a-x,y)$$

In this case to reflex over $x=-1$ we shift $xmapsto x+1$, reflect $mapsto -1-x$ and shift back $mapsto -2-x$

$$(x,y)mapsto (-2-x, y)$$

Similarly for reflecting over horizontal lines, such as $y=b$ involves $(x,y)mapsto (x, 2b-y)$.

Reflecting over a diagonal line is only a bit more complicated, as it involves rotation of the frame of reference.

The practical approach to figure out something like this is to try out various points and see if you can identify a pattern. I think it should be obvious that the $y$ values shouldn't change, so let's just look at the way that points on the $x$-axis reflect to get a sense of the overall pattern.

$$(-2,0) mapsto (0,0)$$
$$(-1,0) mapsto (-1,0)$$
$$(0,0) mapsto (-2,0)$$
$$(1,0) mapsto (-3,0)$$
$$(2,0) mapsto (-4,0)$$

Thus, one could summarize the general pattern as $(x,y) mapsto (-2-x,y)$.