A) (1, -1)
B) (1, 1)
C) (2, 1)
D) (2, 2)
Correct Answer: B
Solution :
Let \[P({{x}_{1}},{{y}_{1}}),\]then the equation of line passing through P and whose gradient is m, is \[y-{{y}_{1}}=m(x-{{x}_{1}})\] Now according to the condition \[\frac{-2m+(m{{x}_{1}}-{{y}_{1}})}{\sqrt{1+{{m}^{2}}}}+\frac{2+(m{{x}_{1}}-{{y}_{1}})}{\sqrt{1+{{m}^{2}}}}+\frac{1-m+(m{{x}_{1}}-{{y}_{1}})}{\sqrt{1+{{m}^{2}}}}=0\] Þ \[3-3m+3m{{x}_{1}}-3{{y}_{1}}=0\Rightarrow {{y}_{1}}-1=m({{x}_{1}}-1)\] Since it is a variable line, so hold for every value of m. Therefore \[{{y}_{1}}=1,{{x}_{1}}=1\Rightarrow P(1,\,1)\].
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