Axis of reflection

Question

Is there any way to get the equation of axis of reflection given two intersecting lines without sketching?

Example question: The image of the line $p:\; y-2x=3$ is the line $q:\;2y-x=9$. Find the equation of axis of reflection?

Answer 1

Say $\ell$ is that line (notice that we have actually 2 such lines). Then $\ell$ is angle bisector for that twp lines. Now point $T(x,y)$ is on this bisector iff $d(T,p)=d(T,q)$.

So in your case $${|-2x+y-3|\over \sqrt{5}} = {|-x+2y-9|\over \sqrt{5}}$$

Now solve this equation and you will get (two) solution.

Answer 2

Rewrite your two lines $p,q$ in parametric form with unit speed as

$$p(t) = \frac1{\sqrt{5}}(1,2)t+(1,5), \quad q(t) = \frac1{\sqrt{5}}(2,1)t+(1,5)$$ for $t \in \Bbb{R}$ with $S =(1,5)$ being their intersection point.

Now for any $t \in \Bbb{R}$ the points $p(t)$ and $q(\pm t)$ are equally far from $S$ since $$|S-p(t)| = \left|\frac1{\sqrt{5}}(1,2)(t)\right| =|t| = \left|\frac1{\sqrt{5}}(2,1)(\pm t)\right| = |S-q(\pm t)|.$$ Therefore if the $r_1,r_2$ are two reflexion lines, we have that $r_1$ and $r_2$ contain the midpoints of the points, say, $p(t)$ and $q(\pm t)$ so we can calculate them as $$r_1(t) = \frac{p(t)+q(t)}2 = \frac3{2\sqrt{5}}(1,1)t+(1,5) \implies x-y+4=0,$$ $$r_2(t) = \frac{p(t)+q(-t)}2 = \frac1{\sqrt{5}}(-1,1)t+(1,5) \implies x-y-6=0.$$

Answer 3

I use the given numerical values to demonstrate the logic.

1. Find X(h, k), the point of intersection of p and q. X = (1, 5).

2.1) Let Q be the point that q cuts the x-axis. $Q = ... = (-9, 0)$.

2.2) Similarly, from p, we have $P = ... (-1.5, 0)$

3. Find XQ and XP. Note that $\dfrac {XQ}{XP} = 2 : 1$ (an accurate approimate).

4. Let R = (?, 0) be the point that the required line cut the x-axis. By angle bisector theorem, $\dfrac {XQ}{XP} = \dfrac {QR}{RP} = \dfrac {2}{1}$

4.1) Note that QR + RP = QP = 7.5. Then, R = ... = (-4, 0).

5. Obtain the required by two point form.