1.

If the points (x, 1), (1, 2) and (0, y + 1) are collinear show that \(\frac{1}{x}\) + \(\frac{1}{y}\) = 1.

Answer»

There are two ways to prove it. 

1st way: Area of triangle formed by the given points = 0 if they are collinear.

∴ \(\frac{1}{2}\) [\(x\)(2 – (y + 1) ) + 1((y + 1) – 1) + 0(1 – 2)] = 0

⇒ \(\frac{1}{2}\) [2\(x\) – \(x\)y – \(x\) + y + 0] = 0 

⇒ \(x\) + y – \(x\)y = 0 

⇒ \(x\) + y = \(x\)y 

⇒ \(\frac{x}{xy}\) + \(\frac{y}{xy}\) = 1 ⇒ \(\frac{1}{y}\) + \(\frac{1}{x}\) = 1.

2nd way: Slope of the lines formed by joining these points are equal. 

Slope of line joining (\(x\), 1), (1, 2) = \(\frac{(2-1)}{1-x}\)

Slope of line joining (1, 2), (0, y + 1) = \(\frac{y+1-2}{0-1}\)

⇒ \(\frac{1}{1-x}\) = \(\frac{y-1}{-1}\) ⇒ -1 = (1 – x) (y – 1) 

⇒ –1 = y – \(x\)y – 1 + \(x\) 

⇒ x + y = \(x\)y ⇒  \(\frac{1}{x}\) + \(\frac{1}{y}\) = 1.


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