A) (0, -3)
B) (0, -3/2)
C) (0, 3/2)
D) (0, 3)
Correct Answer: B
Solution :
Point on y-axis is (0, b), therefore by given condition \[{{(0-3)}^{2}}+{{(b-2)}^{2}}={{(0+1)}^{2}}+{{(b-3)}^{2}}\] \[\Rightarrow \,\,9+{{b}^{2}}+4-4b=1+{{b}^{2}}+9-6b\,\,\Rightarrow \,\,b=-\frac{3}{2}\] Hence the point is\[\left( 0,\,-\frac{3}{2} \right)\].
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