A) \[x+z=0\]
B) \[x-3z=0\]
C) \[x-z=0\]
D) \[x-2z=0\]
Correct Answer: D
Solution :
[d] Let \[(h,\,\,k,\,\,\ell )\] be the point which is equidistant from the points (1, 2, 3) and (3, 2, -1) \[\Rightarrow \sqrt{{{(h-1)}^{2}}+{{(k-2)}^{2}}+{{(\ell -3)}^{2}}}\] \[\Rightarrow \sqrt{{{(h-3)}^{2}}+{{(k-2)}^{2}}+{{(\ell +1)}^{2}}}\] \[\Rightarrow \,\,\,\,{{(h-1)}^{2}}+{{(\ell -3)}^{2}}={{(h-3)}^{2}}+{{(\ell +1)}^{2}}\] \[\Rightarrow {{h}^{2}}+1-2h+{{\ell }^{2}}-6\ell +9={{h}^{2}}-6h+9+{{\ell }^{2}}\] \[+2\ell +1\] \[\Rightarrow -2h-6\ell =-6h+2\ell \Rightarrow 6h-2h-6\ell -2\ell =0\] \[\Rightarrow 4h-8\ell =0\Rightarrow h-2\ell =0\] Putting h = x and \[\ell =z\] We get locus of points \[(h,k,\ell )\] as, \[x-2z=0\]
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