| If a point \[(x,y)\]in \[xy\]-plane is equidistant from \[(-1,1)\]and \[(4,3)\]then |
A) \[10x+4y=23\]
B) \[6x+4y=23\]
C) \[-x+y=7\]
D) \[4x+3y=0\]
Correct Answer: A
Solution :
| Distance between \[(x,y)\]and \[(-1,1)\] |
| \[=\sqrt{{{(y-1)}^{2}}+{{(x+1)}^{2}}}\] |
| Distance between \[(x,y)\]and \[(4,3)\] |
| \[=\sqrt{{{(y-3)}^{2}}+{{(x-4)}^{2}}}\] |
| \[\because \]Point are equidistant |
| \[=\sqrt{{{(y-1)}^{2}}+{{(x+1)}^{2}}}=\sqrt{{{(y-3)}^{2}}+{{(x-4)}^{2}}}\] |
| On squaring both side, we get |
| \[{{(y-1)}^{2}}+{{(x+1)}^{2}}={{(y-3)}^{2}}+{{(x-4)}^{2}}\] |
| \[\Rightarrow \] \[{{y}^{2}}+1-2y+{{x}^{2}}+1+2x\] |
| \[={{y}^{2}}+9-6y+{{x}^{2}}+16-8x\] |
| \[\Rightarrow \] \[2x-2y+2=-\,\,6y-8x+25\] |
| \[\therefore \] \[10x+4y=23\] |
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