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 $\dfrac{{|A{x_1} + B{y_1} + C|}}{{\sqrt {{A^2} + {B^2}} }}$ .