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

Let the vertex of triangle be A(x, y). Then the vertex A(x, y) is equidistant from B and C because ABC is an isosceles triangle. Therefore,
$(\mathrm{x}-1{)}^2+(\mathrm{y}-3{)}^2=(\mathrm{x}+2{)}^2+(\mathrm{y}-7{)}^2$ or $6\mathrm{x}-8\mathrm{y}+43=0$
Thus, any point lying on this line can be the vertex A except the midpoint $(-1/2,5)$ of BC. Hence, vertex A is (5/6, 6).