Question

The algebraic sum of the perpendicular distances from the points $A(-2,0),B(0,2)$ and $C(1,1)$ to a variable line be zero, then all such lines

• A. Are parallel
• B. Passes through a fixed point $(0,0)$
• C. From a square
• D. Passes through the centroid of $\Delta ABC$.

Answer 1

Answer: (D)

Passes through the centroid of $\Delta ABC$.

$y=mx+c$
$y-mx-c=0$
$\frac{0-m\times (-2)-c}{\sqrt{1+m^2}}+\frac{2-0-c}{\sqrt{1+m^2}}+\frac{1-m-c}{\sqrt{1+m^2}}=0$
$2m-c+2-c+1-m-c=0$
$m+3-3c=0$ $........\left(1\right)$

$y=mx+c$

From equation (1)

$1=m\left(\frac{-1}{3}\right)+c$

This line passes through $(\frac{-1}{3},1)$
Centroid of the triangle $=(\frac{-2+0+1}{3},\frac{0+2+1}{3})$
$=(\frac{-1}{3},1)$
Checking for option $D$ , we find that it is the centroid of $△ABC$