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A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point. \[\] [ Hint : $\large\frac{x}{a}$$ + \large\frac{y}{b}$$=1$ where $ \large\frac{1}{a}$$+\large\frac{1}{b}$ = constant = $ \large\frac{1}{k}$ (say ). This implies that $ \large\frac{k}{a}$$+\large\frac{k}{b}$$=1 \Rightarrow $ line passes through the fixed point (k, k).]

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1 Answer

  • Equation of a line in its intercept form is $ \large\frac{x}{a}$$+\large\frac{y}{b}$$=1$
Step 1 :
Let the equation of the line be $ \large\frac{x}{a}$$+\large\frac{y}{b}$$=1$
Its intercepts on the $x$ and $y$ axes are $a$ and $b$ respectively.
It is given that
$ \large\frac{1}{a}$$+\large\frac{1}{b}$ = constant = $k$ ( assume )
$ \therefore \large\frac{1}{ka}$$+ \large\frac{1}{kb}$$=1$
According to the given condition :
$ \large\frac{\Large\frac{1}{k}}{a}$$+\large\frac{\Large\frac{1}{k}}{b}$$=1$
Hence $ \large\frac{1}{k}$$,\large\frac{1}{k}$ satisfies the equation.
$ \large\frac{x}{a}$$+\large\frac{y}{b}$$=1$
Hence the line passes through the fixed point. $ \bigg( \large\frac{1}{k}$$, \large\frac{1}{k} \bigg)$
answered Jul 1, 2014 by thanvigandhi_1

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