A variable line passes through a fixed point P.

Solution:

Let the variable line be $a x+b y=1$

We know that, length of the perpendicular from $(p, q)$ to the line $a x+b y+c=0$

is

$\mathrm{d}=\left|\frac{\mathrm{ap}+\mathrm{bq}+\mathrm{c}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|$

$\mathrm{d}_{1}=\left|\frac{2 \times \mathrm{a}+0 \times \mathrm{b}-1}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|$

$=\frac{2 a-1}{\sqrt{a^{2}+b^{2}}}$

Now, perpendicular distance from B $(0,2)$

$\mathrm{d}_{2}=\left|\frac{0 \times \mathrm{a}+2 \times \mathrm{b}-1}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|$

$=\frac{2 b-1}{\sqrt{a^{2}+b^{2}}}$

Now, perpendicular distance from $C(1,1)$

$\mathrm{d}_{3}=\left|\frac{1 \times \mathrm{a}+1 \times \mathrm{b}-1}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|$

$=\frac{a+b-1}{\sqrt{a^{2}+b^{2}}}$

It is given that the algebraic sum of the perpendicular from the given points $(2,$, $0),(0,2)$ and $(1,1)$ to this line is zero.

$d_{1}+d_{2}+d_{3}=0$

Substituting the values we get

$\therefore \frac{2 a-1}{\sqrt{a^{2}+b^{2}}}+\frac{2 b-1}{\sqrt{a^{2}+b^{2}}}+\frac{a+b-1}{\sqrt{a^{2}+b^{2}}}=0$

⇒ 2a – 1 + 2b – 1 + a + b – 1 = 0

⇒ 3a + 3b – 3 = 0

⇒ a + b – 1 = 0

⇒ a + b = 1

So, the equation ax + by = 1 represents a family of straight lines passing through a fixed point.

Comparing the equation ax + by = 1 and a + b = 1, we get

x = 1 and y = 1

So, the coordinates of fixed point is (1, 1)