Points (a, 0), (0, b) and (1, 1) are collinear, if :

Engineering

Mathematics

Locus of a Point

Various Form of a Straight Line

Question

Points (a, 0), (0, b) and (1, 1) are collinear, if :

$\frac{1}{a} + \frac{1}{b} = 1$

$\frac{1}{a} - \frac{1}{b} = 1$

a + b = 1

a – b = ab

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Step 1 : Calculating

The given points are (a, 0), (0, b), (1, 1)

⇒ Slope of line connecting (a, 0) and (0, b) = Slope of line connecting (0, b) and (1, 1)

Step 2 : Calculating

Slope

= \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

Consider the points (a, 0) and (0, b)

⇒ Slope

= \frac{0 - b}{a - 0} = \frac{- b}{a}

Consider the points (0, b) and (1, 1)

⇒ Slope

= \frac{b - 1}{0 - 1} = - (b - 1)

⟹ \frac{- b}{a} = - (b - 1)

⇒ ab – a = b

⇒ a + b = ab

⟹ \frac{1}{a} + \frac{1}{b} = 1

Hence, The pxression

\frac{1}{a} + \frac{1}{b} = 1

is true for the given conditions.