<s>[INST] Equation of line passing through point. The straight line $3x + 4y + 5 = 0 $ and $4x - 3y - 10 = 0$ intersect at point $A$. Point $B$ on line $3x + 4y + 5 = 0 $ and point C on line $4x - 3y - 10 = 0$ are such that $d(A,B)=d(A,C)$. 
Find the equation of line passing through line $\overline{BC}$ if it also passes through point $(1,2)$.
I have found out slope of both the lines through $A$: $\frac{-3}{4}$ and $\frac{4}{3}$.
I can't figure out how to solve it.
 [/INST] inspection shows the lines are perpendicular and meet at $(-1,2)$.
define variables $(w,z)$ by:
$$
\begin{pmatrix} w \\ z \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 4 & -3 \end{pmatrix}
\left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \end{pmatrix}\right)
$$
the equations of the lines are now $w=0$ and $z=0$, so the lines perpendicular to their bisector have equation $w+z=k$ for some $k$, i.e.
$$
\begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} w \\ z \end{pmatrix} = k
$$
in the original coordinates the line is
$$
\begin{pmatrix} 1 & 1 \end{pmatrix}\begin{pmatrix} 3 & 4 \\ 4 & -3 \end{pmatrix}
\left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \end{pmatrix}\right) = k
$$
or
$$
7x+y =k'
$$
and since the line passes through $(1,2)$ we must have $k'=9$
</s>