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Question

For non-negative real numbers h1, h2, h3, k1, k2, k3, if the algebraic sum of the perpendiculars drawn from the points (2,k1), (3,k2), (7,k3), (h1,4), (h2,5), (h3,3) on a variable line passing through (2,1) is zero, then

A
h1=1,h2=1,h3=2
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B
k1=k2=k3=0
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C
h1=h2=h3=0
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D
k1=1,k2=2,k3=3
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Solution

The correct option is C h1=h2=h3=0
Let the equation of the variable line be ax+by+c=0. It is given that
6i=1axi+byi+ca2+b2=0(algebraic sum of perpendiculars from given six points is 0)6i=1(axi+byj+c)=0
6i=1axi+6i=1byi+6i=1c=06i=1axi+6i=1byi+6c=0
axi6+byi6+c=0

Hence, the line ax+by+c=0 always passes through point xi6,yi6
But it is given line passes through point (2,1)
Hence, 2+3+7+h1+h2+h36=2
h1+h2+h3=0
h1=0, h2=0, h3=0
(as h1, h2, h3 are non-negative).

Similarly, k1+k2+k3+4+536=1
k1=k2=k3=0

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