If the length of the perpendicular from the point (1, 1) to the line a

1.	If the length of the perpendicular from the point (1, 1) to the line ax – by + c = 0 be unity, Show that \(\frac{1}{c} + \frac{1}{a} - \frac{1}{b} = \frac{c}{2ab}.\)
Answer» Given: Line ax – by + c = 0 and point (1, 1) To prove: \(\frac{1}{c} + \frac{1}{a} - \frac{1}{b} = \frac{c}{2ab}\) Concept Used: Distance of a point from a line. Explanation: The distance of the point (1, 1) from the straight line ax − by + c = 0 is 1 ∴ 1 = \(\|\frac{a-b+c}{\sqrt{a^2+b^2}}\|\) ⇒ a² + b² + c²– 2ab + 2ac – 2bc = a² + b² ⇒ ab + bc – ac = \(\frac{c^2}{2}\) Dividing both the sides by abc, we get: \(\frac{1}{c} + \frac{1}{a} - \frac{1}{b} = \frac{c}{2ab}\) Hence proved.

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

Answer»

Given:

Line ax – by + c = 0 and point (1, 1)

To prove:

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

Concept Used:

Distance of a point from a line.

Explanation:

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

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

⇒ a² + b² + c²– 2ab + 2ac – 2bc = a² + b²

⇒ ab + bc – ac = \(\frac{c^2}{2}\)

Dividing both the sides by abc, we get:

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

Hence proved.