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Q.
If algebraic sum of distances of a variable line from points $ (2,0) $ , $ (0,2) $ and $ (-2- 2) $ is zero, then the line passes through the fixed point
AMUAMU 2011Straight Lines
Solution:
Let the variable line be $a x+b y+c=0$ Given, the algebraic sum of the perpendicular from the points $(2,0),(0,2)$ and $(1,1)$ to this line is zero $\therefore \left|\frac{2 \times a + b \times 0 + c }{\sqrt{ a ^{2}+ b ^{2}}}\right|+\left|\frac{ a \times 0 + b \times 2 + c }{\sqrt{ a ^{2}+ b ^{2}}}\right|+\left|\frac{ a \times 1 + b \times 1 + c }{\sqrt{ a ^{2}+ b ^{2}}}\right|= 0$
$\Rightarrow \pm\left(\frac{2 a + c }{\sqrt{ a ^{2}+ b ^{2}}}\right) \pm\left(\frac{2 b + c }{\sqrt{ a ^{2}+ b ^{2}}}\right) \pm\left(\frac{ a + b + c }{\sqrt{ a ^{2}+ b ^{2}}}\right)=0$
$\Rightarrow 2 a + c + 2 b + c + a + b + c = 0$
$\Rightarrow 3 a +3 b +3 c = 0$
$\Rightarrow a + b + c = 0$
This is a linear relation between $a , b$ and $c$. So, the equation $a x+b y+c=0$ represents a family of straight line passing through a fixed point. Comparing $a x+b y+c=0$ and $a+b+c=0$ We obtain The coordinates of fixed point are $(1,1)$.