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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 coordinates of the point P \[\] [Hint: Let the slope of the line be $m$. Then the equation of the line passing through the fixed point $P (x_1 , y_1 )$ is $y - y_1 = m (x - x_1 )$. Taking the algebraic sum of perpendicular distances equal to zero, we get $ y - 1 = m (x - 1). $ Thus $(x_1 , y_1 ) $ is $(1, 1)$.]

$\begin {array} {1 1} (A)\;(0,0) & \quad (B)\;(1,1) \\ (C)\;(-1, -1) & \quad (D)\;\text{None of the above} \end {array}$

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  • Length of the perpendicular from $(x_1, y_1)$ to the line $ax+by+c=0$ is $ p = \bigg| \large\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \bigg|$
Step 1 :
Let the variable line be $ ax+by=1$
It is given that the algebraic sum of the perpendicular from the points $(2,0), (0,2 ), (1,1)$ to this line is zero.
$ \therefore \large\frac{(2)a+(0)b-1}{\sqrt{a^2+b^2}}$$+ \large\frac{(0)(a)+(2)b-1}{\sqrt{a^2+b^2}}$$+ \large\frac{a+b-1}{\sqrt{a^2+b^2}}$$=0$
$ \Rightarrow 3a+3b-3=0$
$ \Rightarrow a+b-1=0$
$\Rightarrow a +b=1$
This is the linear relation between the parameter $a$ and $b$.
So the equation $ax+by=1$ represents a family of straight lines passing through a fixed point.
Comparing $ax+by=1$ and $a+b=1$
We obtain that the coordinates of the fixed point are (1,1)
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answered Jul 1, 2014 by thanvigandhi_1

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