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Question

# Let the algebraic sum of the perpendicular distance from the points (2,0), (0,2) and (1,1) to a variable straight line be zero; then the line passes through a fixed point whose coordinates are:

A
(1,1)
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B
(12,12)
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C
(32,32)
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D
centroid of triangle
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Solution

## The correct option is A (1,1)We have, (x1,y1)=(2,0) (x2,y2)=(0,2) (x3,y3)=(1,1) Then, We know that, Equation of line ax+by+c=0 Distance from point (2,0) D1=(2a+0b+c)k......(1) Distance from point (0,2) D2=(0a+2b+c)k......(2) Distance from point (1,1) D3=(a+b+c)k......(3) Where k=√a2+b2 Now, on adding (1)+(2)+(3) to, and we get, D1+D2+D3=3a+3b+3ck D1+D2+D3=3(a+b+c)k As the sum is zero. Hence, a+b+c=0 From equation of line, ax+by+c=0 Taking (x1,y1)=(1,1) a+b+c=0 Hence, this is the answer.

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