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

Solution: Let the vertex of triangle be A(x,y) . Then the vertex A(x,y) is Equdistant from B and C because ABC is an  Isosceles triangle,

Therefore,

$\\ (x-1)^2+(y-3)^2=(x+2)^2+(y-7)^2\\ \\\Rightarrow 6x-8y+43=0$

Thus ,any point lying on this line can be the vertex A except the mid point $\\ (-1/2,5)$ of BC .

Hence vertex A is $(5/6,6)$

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