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# A variable line passes through a fixed point P.

Question:

A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero. Find the coordinates of the 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)