Triangle and points

Question

Given an isosceles triangle ABC:
$ AC = AB $
$ B = (1, 2) $
$ C = (6, 7) $
$ y(AC) = \frac{x}{3} + 5 $
$ y(BC) = x + 1 $

How do find $A$? I know this must be something very simple that I'm missing and I'm absolutely embarrassed.

Answer 1

Hint:

Since $A$ belongs to the line $AC$, let $(a,\frac{a}{3}+5)$ be their coordinates. \begin{align} AC=AB\quad \iff \quad \sqrt{\left(6-a\right)^2+\left(7-\frac{a}{3}-5\right)^2}&=\sqrt{\left(1-a\right)^2+\left(2-\frac{a}{3}-5\right)^2} \end{align}