Question

Let the algebraic sum of the perpendicular distances from the points (2, 0), (0,2) and (1, 1) to a variable straight line be zero, then the line passes through a fixed point whose coordinates are
A.{1, 1}
B.{2, 3}
C. $ \left( {\dfrac{3}{5},\dfrac{3}{5}} \right) $
D.None of these

Answer 1

Hint: In this question, we have given an algebraic sum of the perpendicular distances from the points to a variable straight line is zero; Here the variable straight line is represented by ax + by + c = 0. Then we will try to find the perpendicular distance from each point to the variable straight-line ax + by + c = 0.
Distance of point from a line, d = $ \dfrac{{|A{x_1} + B{y_1} + C|}}{{\sqrt {{A^2} + {B^2}} }} $

Complete step-by-step answer:
Let variable straight line be ax + by + c = 0 ………… (1)
Perpendicular distance ( $ {P_1} $ ) from point (2, 0) to ax + by + c = 0 is = $ \dfrac{{2a + c}}{{\sqrt {{a^2} + {b^2}} }} $
Perpendicular distance ( $ {P_2} $ ) from point (0, 2) to ax + by + c = 0 is = $ \dfrac{{2b + c}}{{\sqrt {{a^2} + {b^2}} }} $
Perpendicular distance ( $ {P_3} $ ) from point (1, 1) to ax + by + c = 0 is = $ \dfrac{{a + b + c}}{{\sqrt {{a^2} + {b^2}} }} $
Now, adding $ {P_1} $ , $ {P_2} $ , and $ {P_3} $ , we get;
  $ {P_1} + {P_2} + {P_3} = 0 $
 $ \Rightarrow \dfrac{{2a + c}}{{\sqrt {{a^2} + {b^2}} }}\, + \,\dfrac{{2b + c}}{{\sqrt {{a^2} + {b^2}} }}\, + \,\dfrac{{a + b + c}}{{\sqrt {{a^2} + {b^2}} }}\, = \,0 $
 $ \Rightarrow $ 2a + c + 2b + c + a + b + c = 0
 $ \Rightarrow $ 2a + a + 2b + b + 3c = 0
 $ \Rightarrow $ 3a + 3b + 3c = 0
 $ \Rightarrow $ a + b + c = 0 …………. (2)
Comparing equation (1) and (2), we get x = 1 and y = 1.
Hence the line passes through a fixed point whose coordinates are (1, 1).
So, the correct answer is “Option A”.

Note: You have to be careful while solving this question. You have to find out the perpendicular distance from each point to that straight-line equation using the formula of $ \dfrac{{|A{x_1} + B{y_1} + C|}}{{\sqrt {{A^2} + {B^2}} }} $ .