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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 y−y1=m(x−x1). Now according to the condition −2m+(mx1−y1)√1+m2+2+(mx1−y1)√1+m2+1−m+(mx1−y1)√1+m2=0⇒3−3m+3mx1−3y1=0⇒y1−1=m(x1−1)Since it is a variable line, so hold for every value of m. Therefore y1=1,x1=1⇒P(1,1)

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