• # question_answer A variable line passes through a fixed point P. The algebraic sum of the perpendicular drawn from (2,0), (0, 2) and        (1, 1) on the line is zero, then the coordinates of the P are [IIT 1991; AMU 2005]                                        A)            (1, -1)                                         B)            (1, 1) C)            (2, 1)                                          D)            (2, 2)

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 ${{y}_{1}}=1,{{x}_{1}}=1\Rightarrow P(1,\,1)$.