Question

If algebraic sum of distances of a variable line from points $(2,0)$ , $(0,2)$ and $(-2- 2)$ is zero, then the line passes through the fixed point

• A. $(-1,-1)$
• B. $(1,1)$
• C. $(2,2)$
• D. $(0,0)$

Answer 1

Answer: (B)

$(1,1)$

Answer 2

Let the variable line be $a x+b y+c=0$ Given, the algebraic sum of the perpendicular from the points $(2,0),(0,2)$ and $(1,1)$ to this line is zero $\therefore\left|\frac{2 \times a + b \times 0 + c }{\sqrt{ a ^{2}+ b ^{2}}}\right|+\left|\frac{ a \times 0 + b \times 2 + c }{\sqrt{ a ^{2}+ b ^{2}}}\right|+\left|\frac{ a \times 1 + b \times 1 + c }{\sqrt{ a ^{2}+ b ^{2}}}\right|= 0$
$\Rightarrow \pm\left(\frac{2 a + c }{\sqrt{ a ^{2}+ b ^{2}}}\right) \pm\left(\frac{2 b + c }{\sqrt{ a ^{2}+ b ^{2}}}\right) \pm\left(\frac{ a + b + c }{\sqrt{ a ^{2}+ b ^{2}}}\right)=0$
$\Rightarrow 2 a + c + 2 b + c + a + b + c = 0$
$\Rightarrow 3 a +3 b +3 c = 0$
$\Rightarrow a + b + c = 0$
This is a linear relation between $a , b$ and $c$. So, the equation $a x+b y+c=0$ represents a family of straight line passing through a fixed point. Comparing $a x+b y+c=0$ and $a+b+c=0$ We obtain The coordinates of fixed point are $(1,1)$.