Axis of reflection

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.

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 $$r_1(t) = frac{p(t)+q(t)}2 = frac3{2sqrt{5}}(1,1)t+(1,5) implies x-y+4=0,$$ $$r_2(t) = frac{p(t)+q(-t)}2 = frac1{sqrt{5}}(-1,1)t+(1,5) implies x-y-6=0.$$

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.