The quadratic equation whose roots are the x and yintercepts of the...

Equation of the line making intercepts a and $b$ on the axes is $\frac{x}{a}+\frac{y}{b}=1$ Since, it passes through (1,1) $\Rightarrow \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{ b}}=1$ Also the area of the triangle formed by the line and the axes is $\mathrm{A}$. $\therefore \frac{1}{2}$\mathrm{ab}=\mathrm{A}$\Rightarrow \mathrm{ab}=2 \mathrm{ A}$ From eqs. (i) and (ii), we get, $a+b=2 A$ Hence, a and b are the roots of the eq. $x^{2}-(a+b) x+a b=0 \Rightarrow x^{2}-2 A x+2 A=0$