I am having trouble with a coordinate geometry question:
Triangle ABC has been reflected. The pre-image has vertices at
A(2, -1), B(4, 0), and C(3, -3). The image has a vertex A’ at (-1, 0).
Where are B’ and C’ located and what is the equation for the line
of reflection?
I know how to find the line of reflection, but to find the midpoints you would need more than one point. Is guess-and-check the only method to solving this, or is there a formula that will help?
Best Answer
We start with the perpendicular bisector of $\overline{AA'}$, as per stewbasic’s comment. This line passes through the midpoint $M=(A+A')/2 = (1/2,-1/2)$ and has a the vector $\mathbf n=A-A'=(3,-1)$ for a normal, so an equation for this line is $\mathbf n\cdot\mathbf x=\mathbf n\cdot\mathbf M$, or $3x+y=2$.
We can now find the images of the other two points using the formula for reflection in a line through the origin with normal $\mathbf n$: $$\mathbf v'= \mathbf v-2{\mathbf n\cdot\mathbf v\over\mathbf n\cdot\mathbf n}\mathbf n.$$ Our line of reflection doesn’t pass through the origin, though, so we’ll have to do a couple of translations to get the right formula. The reflection formula for our line is thus $$\begin{align}\mathbf v' &= (\mathbf v-M)-2{\mathbf n\cdot(\mathbf v-M)\over\mathbf n\cdot\mathbf n}\mathbf n+M\\ &= \mathbf v-2{\mathbf n\cdot(\mathbf v-M)\over\mathbf n\cdot\mathbf n}\mathbf n.\end{align}$$