A) \[\left( \frac{11}{2},0 \right)\]
B) \[(5,0)\]
C) \[(6,0)\]
D) \[(5,3)\]
Correct Answer: A
Solution :
| Let point on X-axis be \[P(x,0)\] |
| Then\[{{(QP)}^{2}}={{(PR)}^{2}}\] |
| \[\therefore \,\,\,\,\,\,\,{{(x-2)}^{2}}+{{(0-3)}^{2}}={{(9-x)}^{2}}+{{(3-0)}^{2}}\] |
| \[\Rightarrow \,\,{{x}^{2}}+4-4x+9=81+{{x}^{2}}-18x+9\] |
| \[\Rightarrow \,\,\,\,14x=77\,\,\,\,\Rightarrow \,\,\,x=\frac{11}{2}\] |
| \[\therefore \] Point on X- axis \[\left( \frac{11}{2},0 \right)\], |
| So, option [a] is correct. |
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