The quadratic equation whose roots are the $\mathrm{x}$ and yintercepts of the line passing through (1,1)…
The quadratic equation whose roots are the $\mathrm{x}$ and yintercepts of the line passing through (1,1) and making a triangle of area $\mathrm{A}$ with the co-ordinate axes is
$x^{2}+A x+2 A=0$
$x^{2}-2 A x+2 A=0$
$x^{2}-A x+2 A=0$
None of these
Solution
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. $x^{2}-(a+b) x+a b=0 \Rightarrow x^{2}-2 A x+2 A=0$