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Question

A variable line passes through a fixed point P, The algebraic sum of the perpendicular distances from (2, 0), (0, 2) and (1, 1) to the line is zero, then the coordinates of the P are

A
(1, -1)
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B
(1, 1)
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C
(2, 1)
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D
(2, 2)
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Solution

The correct option is B (1, 1)
Let P(x1,y1), then the equation of line passing through P and whose gradient is m, is yy1=m(xx1). Now according to the condition
2m+(mx1y1)1+m2+2+(mx1y1)1+m2+1m+(mx1y1)1+m2=033m+3mx13y1=0y11=m(x11)Since it is a variable line, so hold for every value of m. Therefore y1=1,x1=1P(1,1)

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