This determines the gradient of the required line. Use B\((6,3)\) and \(m\). The perpendicular bisector passes through the midpoint of AC. This determines the point on the required line.

In the diagram below, AB is the chord of a circle with centre O. OM is perpendicular to AB (meet at a right angle). Look at triangles OAM and OBM. The hypotenuses (OA and OB) are the same ...