A) \[10x+4\,\,y=23\]
B) \[6x+4\,\,y=23\]
C) \[-x+y=7\]
D) \[4x+3\,\,y=0\]
Correct Answer: A
Solution :
Ans. Hence, the points are at equidistance. Then, \[\sqrt{{{(y-1)}^{2}}+{{(x+1)}^{2}}}=\sqrt{{{(y-3)}^{2}}+{{(x-4)}^{2}}}\] On squaring both side, we get \[y+1-2y\,\,\,\,x\,\,\,\,1+2x\] \[={{y}^{2}}=+\,9-6y+{{x}^{2}}+16-8x\] \[\Rightarrow \] \[2x-2y+2=-6y-8x+25\] \[\Rightarrow \] \[10x+4y=23\]
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