MCQ
If the points $(a,\,0),\;(0,\,b)$ and $(1, 1)$ are collinear, then
  • A
    $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = 1$
  • B
    $\frac{1}{{{a^2}}} - \frac{1}{{{b^2}}} = 1$
  • $\frac{1}{a} + \frac{1}{b} = 1$
  • D
    $\frac{1}{a} - \frac{1}{b} = 1$

Answer

Correct option: C.
$\frac{1}{a} + \frac{1}{b} = 1$
c
(c) $\frac{1}{2}\,\left| {\,\begin{array}{*{20}{c}}a&0&1\\0&b&1\\1&1&1\end{array}\,} \right| = 0\,\,$

$\Rightarrow \,\,a\,(b - 1) + 0 + 1\,( - b) = 0$

$ \Rightarrow \,\,ab - a - b = 0\,\,\,$

$\Rightarrow \,\,\frac{1}{a} + \frac{1}{b} = 1$.

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