MCQ
A variable line passes through a fixed point $P$. The algebraic sum of the perpendicular drawn from $(2,0)$, $(0, 2)$ and $(1, 1)$ on the line is zero, then the coordinates of the $P$ are
  • A
    $(1,\, -1)$
  • $(1, \,1)$
  • C
    $(2, \,1)$
  • D
    $(2,\, 2)$

Answer

Correct option: B.
$(1, \,1)$
b
(b) Let $P({x_1},{y_1}),$ then the equation of line passing through $P$ and whose gradient is $m$, is $y - {y_1} = m(x - {x_1})$

Now according to the condition

$\frac{{ - 2m + (m{x_1} - {y_1})}}{{\sqrt {1 + {m^2}} }} + \frac{{2 + (m{x_1} - {y_1})}}{{\sqrt {1 + {m^2}} }} + \frac{{1 - m + (m{x_1} - {y_1})}}{{\sqrt {1 + {m^2}} }} = 0$

==> $3 - 3m + 3m{x_1} - 3{y_1} = 0 \Rightarrow {y_1} - 1 = m({x_1} - 1)$

Since it is a variable line, so hold for every value of $m$. Therefore ${y_1} = 1,{x_1} = 1 \Rightarrow P(1,\,1)$.

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