If the algebraic sum of the perpendicular distances from the points $(2,0),(0,2)$ and $(1,1)$ to a variable line is zero, then the variable line always passes through a fixed point. The coordinates of that point are

• A

  $(0,0)$
• B

  $(2,0)$
• C

  $(0,2)$
• Correct answer
  D

  $(1,1)$

Solution

now, perpendicular distance from $(2,0),(0,2)$ and $(1,1)$ to the line is respectively $$ \mathrm{d}_1=\frac{2 \mathrm{a}+\mathrm{c}}{2}, \mathrm{~d}_2=\frac{2 \mathrm{~b}+\mathrm{c}}{2} \text { and }_3=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2} $$ Since, $d_1+d_2+d_3=0$ $$ \begin{aligned} & \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=0 \\ & \Rightarrow \mathrm{c}=-\mathrm{a}-\mathrm{b} \end{aligned} $$ Now from equation (i) $$ \begin{aligned} & c=-a x,-b y \\ & \Rightarrow(x, y)=(1,1) \end{aligned} $$

Hence, the correct answer is option: 'D'.