If the points (a, 0), (0, b) and (1, 1) are collinear then which of the following is true ?

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

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

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

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

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Solution

The correct option is A $\frac{1}{a} + \frac{1}{b} = 1$
Given that three points are collinear, $(a, o), (o, b)$ and $(1, 1)$
Since the points are collinear, area of triangle formed by these points is equal to O.
$\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = 0 \frac{1}{2} [a [b - 1] + 0 [1 - 0] + 1 (0 - b)] = 0 \Rightarrow a b - a - b = 0 \Rightarrow a + b = a b ∴ \frac{1}{a} + \frac{1}{b} = 1$

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Similar questions

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

(a, 0), (0, b)

(1, 1)

(a_{1}, b_{1})

(a_{2}, b_{2})

(a_{1} + a_{2}, b_{1} + b_{2})

a_{1} b_{2} = a_{2} b_{1}

(a, 0), (0, b)

(1, 1)

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

If the lines a_{1} x + b_{1} y + 1 = 0, a_{2} x + b_{2} y + 1 = 0 and a_{3} x + b_{3} y + 1 = 0 are concurrent, then the points (a_{1}, b_{1}), (a_{2}, b_{2}) and (a_{3}, b_{3}) will be collinear .

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