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

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Solution

The distance of the point (1, 1) from the straight line ax − by + c = 0 is 1

$∴ 1 = |\frac{a - b + c}{\sqrt{a^{2} + b^{2}}}| \Rightarrow a^{2} + b^{2} + c^{2} - 2 a b + 2 a c - 2 b c = a^{2} + b^{2} \Rightarrow a b + b c - a c = \frac{c^{2}}{2}$

Dividing both the sides by abc, we get:

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

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

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. Let

a, b, c

be such that

b (a + c) \neq 0

∣ ∣ ∣ ∣ \begin{matrix} a & a + 1 & a - 1 - b & b + 1 & b - 1 c & c - 1 & c + 1 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} a + 1 & b + 1 & c - 1 a - 1 & b - 1 & c + 1 (- 1)^{n + 2} a & (- 1)^{n + 1} b & (- 1)^{n} c \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

, then the value of

n

Q. Let a, b and c be such that b(a+c)

\neq

0.
If

∣ ∣ ∣ ∣ \begin{matrix} a & a + 1 & a - 1 - b & b + 1 & b - 1 c & c - 1 & c + 1 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} a + 1 & b + 1 & c - 1 a - 1 & b - 1 & c + 1 (- 1)^{n + 2} a & (- 1)^{n - 1} b & (- 1)^{n} c \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

, then the value of n is

Let a, b, c be such that (b+c) ≠0. If

Then, the value of 'n' is

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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.

The distance of the point (1, 1) from the straight line ax − by + c = 0 is 1 ∴ 1=a-b+ca2+b2⇒a2+b2+c2-2ab+2ac-2bc=a2+b2⇒ab+bc-ac=c22 Dividing both the sides by abc, we get: 1c+1a-1b=c2ab

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