Question

$ABC$ 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

• A. $(1,6)$
• B. $(-1/2,5)$
• C. $(5/6,6)$
• D. none of these

Answer 1

Answer: (C)

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,
$(x-1{)}^2+(y-3{)}^2=(x+2{)}^2+(y-7{)}^2$
or $6x-8y+43=0$
Thus, any point lying on this line can be the vertex $A$ except the midpoint $(-1/2,5)$ of $BC$.
Hence, vertex $A$ is $(5/6,6)$.