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A variable line passes through a fixed p...

A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2,0), (0,2) and (1,1) on the line is zero. Find the coordinate of the point P.

Text Solution

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Let the equation of variable line be
`ax+by+c=0 " " (1)`
It is given that
`sum_(i=1)^(6)(ax_(i)+by_(i)+c)/(sqrt(a^(2) + b^(2))) = 0, "where" (x_(i), y_(i)), i=1,2,3,4,5,6 " are given points."`
`rArr a((sum_(i=1)^(6)x_(i))/(6)) + b((sum_(i=1)^(6)y_(i))/(6)) +c =0 " " (2)`
Comparing (1) and (2), we get
`x/((sum_(i=1)^(6)x_(i))/(6))=y/((sum_(i=1)^(6)y_(i))/(6))=1`
`therefore (x,y) -= ((sum_(i=1)^(6)x_(i))/(6), (sum_(i=1)^(6)y_(i))/(6))`
This is the fixed point through which the variable (1) passes.
But it is given that the variable line is passing through the point (2,1).
`" So, "(sum_(i=1)^(6)x_(i))/(6) = 2" and " (sum_(i=1)^(6)y_(i))/(6) = 1`
`rArr (2+3+7+h_(1) + h_(2)+h_(3))/(6) =2 " and " (k_(1) + k_(2) + k_(3) +4+5-3)/(6) = 1`
`rArr h_(1) + h_(2)+h_(3) =0 " and " k_(1) + k_(2) + k_(3) = 0`
`" Since " h_(1), h_(2),h_(3), k_(1), k_(2),k_(3) " are non-negative,"h_(1) = h_(2) =h_(3)=0 and k_(1) =k_(2)=k_(3)=0.`
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