If the length of the perpendicular from the point (1, 1) to the line $a x - b y + c = 0$ be unity, show that $\frac{1}{c} + \frac{1}{a} - \frac{1}{b} = \frac{c}{2 a b}$

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Solution

Length of perpendicular from (1, 1) to $a x - b y + c = 0$

$\Rightarrow ∣ ∣ ∣ \frac{a (1) - b (1) + c}{\sqrt{a^{2} + b^{2}}} ∣ ∣ ∣ = 1$

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

$(a - b + c)^{2} = a^{2} + b^{2}$

$a^{2} + b^{2} + c^{2} + 2 a c - 2 b c - 2 a b = a^{2} + b^{2}$

$c^{2} + 2 a c - 2 b c = 2 a b$

$c + 2 a - 2 b = \frac{2 a b}{c}$

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

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

Hence, proved.

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Similar questions

Q. If the lenght of the perpendicular from the point (

1, 1

) to the line ax-by+c=

0

be unity show that

- \frac{1}{c} + \frac{1}{a} + \frac{1}{b} = \frac{c}{2 a b}

Q. If the length of the perpendicular from the point (1, 1) to the line ax − by + c = 0 be unity, show that

\frac{1}{c} + \frac{1}{a} - \frac{1}{b} = \frac{c}{2 a b}

Find the length of the perpendicular from the point $(x^{1}, y^{1})$ to the straight line Ax + By + C = 0, the axes being inclined at an angle $ω$ , and the equation being written such that C is a negative quantity.

Q. I : Length of the perpendicular from

(x_{1}, y_{1})

to the line

a x + b y + c = 0

∣ ∣ ∣ \frac{a x_{1} + b y_{1} + c}{\sqrt{a^{2} + b^{2}}} ∣ ∣ ∣ .

II : The equation of the line passing through

(0, 0)

and perpendicular to

a x + b y + c = 0

b x - a y = 0.

Then which of the following is true?

Q. If

\frac{1}{\sqrt{b} + \sqrt{c}}, \frac{1}{\sqrt{c} + \sqrt{a}}, \frac{1}{\sqrt{a} + \sqrt{b}}

are in A.P., then the family of lines

a x + b y + c = 0

will always passes through the fixed point:

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If the length of the perpendicular from the point (1, 1) to the line ax−by+c=0 be unity, show that 1c+1a−1b=c2ab

Length of perpendicular from (1, 1) to ax−by+c=0 ⇒∣∣∣a(1)−b(1)+c√a2+b2∣∣∣=1 a−b+c=√a2+b2 (a−b+c)2=a2+b2 a2+b2+c2+2ac−2bc−2ab=a2+b2 c2+2ac−2bc=2ab c+2a−2b=2abc c2ab+2a2ab−2b2ab=1c c2ab=1c+1a−1b Hence, proved.

If the length of the perpendicular from the point (1, 1) to the line $a x - b y + c = 0$ be unity, show that $\frac{1}{c} + \frac{1}{a} - \frac{1}{b} = \frac{c}{2 a b}$