Question

$L$ is a variable line such that the algebraic sum of the distances of the points$(1,1),(2,0)$, and $(0,2)$ from the line is equal to zero. The line $L$ will always pass-through

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

Answer 1

The correct option is A

$(1,1)$

Explanation for the correct option:

Step:1 Finding condition of sum of distances of point is zero

Given the points are $(1,1),(2,0)$, and $(0,2)$

Here $(1,1)$ is the mid point of $(0,2)\&(2,0)$

Then the distance form the point $(2,0)\&(1,1)$ is same as the distance from $(0,2)$ to $(1,1)$

Then the algebraic sum of the distance is zero.

Line $L$ is always passes through the point $(1,1)$

Let $y-mx-c=0$ be the equation of line, then we have,

Step:2 Finding slope of the line using algebric sum of distances is zero

The algebraic sum of the distance is zero.

$\Rightarrow \left|\frac{1-m-c}{\sqrt{1+m^2}}\right|+\left|\frac{0+2m-c}{\sqrt{1+m^2}}\right|+\left|\frac{2-c}{\sqrt{1+m^2}}\right|=0$

Taking the denominator to RHS since the denominator is same.

$\Rightarrow \left(1-m-c\right)+\left(2m-c\right)+\left(2-c\right)=0$

Then we get,

$$\begin{gathered} \Rightarrow 2-c=0 \\ \Rightarrow c=2 \end{gathered}$$

Substituting $c=2$ in $1-m-c=0$

$$\begin{gathered} \Rightarrow 1-m-2=0 \\ \Rightarrow -m-1=0 \\ \Rightarrow m=-1 \\ \therefore m=-1,c=2 \end{gathered}$$

Step:3 Finding equation of line and the point

The equation of line becomes

$$\begin{gathered} \Rightarrow y+x-2=0 \\ \Rightarrow y+x=2 \\ \Rightarrow x+y=2 \end{gathered}$$

Therefore the line passes through the point $(1,1)$

Hence, option (A) is the correct answer.