Find the images of (1,0) under reflection in L?

Question

Consider the line

$$L = \{(x,y): x - 2y = 2\}$$

Find the images of $(1,0)$ under reflection in $L$?

Thanks in advance.

Find a parametric representation of the normal from $(1,0)$ towards $L$. Find the parameters that correspond to $(1,0)$ and the intersection with $L$, respectively. The desired point is the same distance on the other side of the intersection. — hmakholm left over Monica, Dec 03 '14 at 13:14

score 0 · Accepted Answer · answered Dec 03 '14 at 13:34

Our straight line is $$L = \{(x,y) : \, y = \frac{1}{2}x - 1\};$$ so the normal to $L$ through the point $(1,0)$ is $$L^\prime = \{ \, (x,y) \, : \, y = -2(x-1) = -2x+2 \}.$$ These two lines intersect in the point $(6/5, -2/5)$.

Now let $(a,b)$ be the image of $(1,0)$ under the reflection in $L$.

Then the point $(6/5, -2/5)$ must be the mid-point of the segment joining $(1,0)$ and $(a,b)$.

Hence we have the equations $$\frac{6}{5} = \frac{1+a}{2},$$ and $$\frac{-2}{5} = \frac{0+b}{2},$$ So $a = 7/5$ and $b = -4/5$.

Thus our required point is $(7/5, -4/5)$.

Thanks for the help, makes much more sense now :) – Joe Dec 03 '14 at 15:27 — Joe, Dec 03 '14 at 15:27

Find the images of (1,0) under reflection in L?

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