Question

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

• A. $(-1,-1)$
• B. $(1,1)$
• C. $(2,2)$
• D. $(0,0)$

Answer 1

Answer: (B)

Let the variable line be $ax+by+c=0$ Given, the algebraic sum of the perpendicular from the points $(2,0),(0,2)$ and $(1,1)$ to this line is zero $\therefore \left|\frac{2\times a+b\times 0+c}{\sqrt{a^2+b^2}}\right|+\left|\frac{a\times 0+b\times 2+c}{\sqrt{a^2+b^2}}\right|+\left|\frac{a\times 1+b\times 1+c}{\sqrt{a^2+b^2}}\right|=0$
$\Rightarrow \pm \left(\frac{2a+c}{\sqrt{a^2+b^2}}\right)\pm \left(\frac{2b+c}{\sqrt{a^2+b^2}}\right)\pm \left(\frac{a+b+c}{\sqrt{a^2+b^2}}\right)=0$
$\Rightarrow 2a+c+2b+c+a+b+c=0$
$\Rightarrow 3a+3b+3c=0$
$\Rightarrow a+b+c=0$
This is a linear relation between $a,b$ and $c$. So, the equation $ax+by+c=0$ represents a family of straight line passing through a fixed point. Comparing $ax+by+c=0$ and $a+b+c=0$ We obtain The coordinates of fixed point are $(1,1)$.