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?
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$\begingroup$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 $$\begin{align}\mathbf v' &= (\mathbf v-M)-2{\mathbf n\cdot(\mathbf v-M)\over\mathbf n\cdot\mathbf n}\mathbf n+M\\ &= \mathbf v-2{\mathbf n\cdot(\mathbf v-M)\over\mathbf n\cdot\mathbf n}\mathbf n.\end{align}$$
$\endgroup$ $\begingroup$As @stewbasic very rightly stated, the line of reflection that sends $A$ to $A'$ is the perpendicular bisector of the line $AA'$ (that is, the line connecting $A$ and $A'$).
The line $AA'$ has gradient $m_1 = \frac{-1 -0}{2-(-1)}= \frac{-1}{3}$. Thus it's perpendicular bisector will be perpendicular to this gradient so will have gradient $m = \frac{-1}{m_1} = 3$. And since it is a bisector, it must pass through the midpoint of the line $AA'$. This midpoint is at: $ x = \frac{(-1)+2}{2} = \frac{-1}{2}$ and $y= \frac{0 + -1}{2} = \frac{-1}{2}$.
So the perpendicular bisector is of the form $y = mx+c$ where $m = 3$ and the line passes through the point $(\frac{-1}{2}, \frac{-1}{2})$. We can thus solve for c giving the line of reflection: $$y = 3x - 2$$
Can you complete the rest from here?
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