# 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}$$=1 \Rightarrow line passes through the fixed point (k, k).] ## 1 Answer Toolbox: • Equation of a line in its intercept form is \large\frac{x}{a}$$+\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)$