Question

A variable line passes through a fixed point P, The algebraic sum of the perpendicular distances from (2, 0), (0, 2) and (1, 1) to the line is zero, then the coordinates of the P are

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

Answer 1

The correct option is B (1, 1)

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
$\frac{-2m+(m{x}_1-y_1)}{\sqrt{1+m^2}}+\frac{2+(m{x}_1-y_1)}{\sqrt{1+m^2}}+\frac{1-m+(m{x}_1-y_1)}{\sqrt{1+m^2}}=0\Rightarrow 3-3m+3m{x}_1-3y_1=0\Rightarrow y_1-1=m(x_1-1)Sinceitisavariableline,soholdforeveryvalueofm.Therefore{y}_1=1,x_1=1\Rightarrow P(1,1)$