A) (0, 7)
B) \[(-7,0)\]
C) \[(0,-7)\]
D) (0, 0)
Correct Answer: B
Solution :
Since, A (x, 0) is equidistant from \[\text{B}\,\,(2-5),\]arc C\[(-\,\,2,9).\] \[\therefore \] \[AB=AC\Rightarrow A{{B}^{2}}=A{{C}^{2}}\] \[\Rightarrow \]\[{{(x-2)}^{2}}+{{(0+5)}^{2}}={{(x+2)}^{2}}+{{(0-9)}^{2}}\] [\[\because \]Distance \[=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}+{{y}_{1}})}^{2}}}\]] \[\Rightarrow \] \[{{x}^{2}}-4x+4+25={{x}^{2}}+4x+4+81\] \[\Rightarrow \] \[-\,\,4x-4x=81-25\] \[\Rightarrow \] \[-\,\,8x=56\] \[\therefore \] \[x=\frac{56}{8}=-7\] So, the point equidistant from given points on the x-axis is \[(-\,\,7,0).\]
You need to login to perform this action.
You will be redirected in
3 sec
