ABC is an isosceles triangle. If the coordinates of the base are. $B\left(1,3\right)$) and, $C\left(-2,7\right)$, the coordinates of vertex A can be 

$$\begin{gathered} Given that the triangle ABC\hspace{0.17em}is Isosceles triangle, A is at equal sides from the vertices B and C \\ So that any point on the perependicular bisector of BC except the midpoint of BC is A \\ The midpoint of BC is \left(-\frac{1}{2},5\right) \\ The perpendicular bsiector of BC\hspace{0.17em}is y-5=\frac{3}{4}\left(x+\frac{1}{2}\right) \\ 4y-20=3x+\frac{3}{2} \\ 6x-8y-43=0 \\ Verify the options to get the vertex A \end{gathered}$$