A) (1, 6)
B) \[\left( -\frac{1}{2},\,5 \right)\]
C) \[\left( \frac{5}{6},\,6 \right)\]
D) None of these
Correct Answer: C
Solution :
The vertex \[A\,(x,\,\,y)\] is equidistant from B and C because ABC is an isosceles triangle. Therefore, \[{{(x-1)}^{2}}+{{(y-3)}^{2}}={{(x+2)}^{2}}+{{(y-7)}^{2}}\] \[\Rightarrow \,\,6x-8y+43=0\] Thus, any point lying on this line can be the vertex A except the mid-point \[\left( -\frac{1}{2},\,\,5 \right)\] of BC.
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