Question

A line passes through $(2,2)$ and has $x$-intercept and $y$-intercept as $\alpha \text{units}$ and $\beta \text{units}$ respectively. It makes a triangle of area $A$ with co-ordinate axes. Then the quadratic equation whose roots are $\alpha$ and $\beta$ is :
($\alpha >0,\beta >0$)

• A. $x^2-2Ax+2A=0$
• B. $x^2-Ax+2A=0$
• C. $x^2+2Ax+2A=0$
• D. $x^2+Ax+2A=0$

Answer 1

The correct option is B $x^2-Ax+2A=0$

Let the equation of line be
$\frac{x}{\alpha }+\frac{y}{\beta }=1$
it passes through $(2,2)$ so
$\frac{2}{\alpha }+\frac{2}{\beta }=1$
$\Rightarrow 2(\alpha +\beta )=\alpha \beta$
Here, $\alpha$ and $\beta$ are intercepts
So the area of triangle $=\frac{1}{2}\alpha \beta =A$
$\Rightarrow \alpha \beta =2A\Rightarrow \alpha +\beta =A$
So, Quadratic equation will be
$x^2-(\alpha +\beta )x+\alpha \beta =0$
$\Rightarrow x^2-Ax+2A=0$