A straight line cuts intercepts from the axes of coordinates the sum of whose reciprocals is a constant. Show that it always passes through a fixed point ?

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Solution

$e q u a t i o n o f l i n e \frac{x}{a} + \frac{y}{b} = 1$
$i n t e r c e p t s w i t h a x i s a a n d b r e s p e c t i v e l y$
$\frac{1}{a} + \frac{1}{b} = c o n s t a n t = \frac{1}{p}$
$\frac{1}{a} + \frac{1}{b} = \frac{1}{p}$
$\frac{P}{a} + \frac{P}{b} = 1$

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A straight line cuts intercepts from the axes of coordinates the sum of whose reciprocals is a constant. Show that it always passes through a fixed point ?

equationoflinexa+yb=1interceptswithaxisaandbrespectively 1a+1b=constant=1p1a+1b=1pPa+Pb=1

$e q u a t i o n o f l i n e \frac{x}{a} + \frac{y}{b} = 1$
$i n t e r c e p t s w i t h a x i s a a n d b r e s p e c t i v e l y$
$\frac{1}{a} + \frac{1}{b} = c o n s t a n t = \frac{1}{p}$
$\frac{1}{a} + \frac{1}{b} = \frac{1}{p}$
$\frac{P}{a} + \frac{P}{b} = 1$