Finding A Quadratic Whose Roots Equals Intercept On Axes and Area Equals A How to find the quadratic equation whose roots are the x and y intercepts of the line passing through $(1,1)$ and making a triangle of area A with the axes? 
Ok I'm getting $(1-m)(1-1/m)=A$ and $(x-(1-m))(x-(1-1/m))=y$.But how to proceed from here?How to get the sum of roots? 
 A: Call the gradient $m$, the x-intercept $a$ and the y-intercept $b$.
Then we have:
$y = mx + b$
$x = \frac{y}{m} + a$
Plugging in our known point (1, 1) and rearranging:
$1 - b = m$
$1 - a =\frac{1}{m}$
As we're not interested in the value of $m$, eliminate it by multiplying the  two equations together:
$(1 - b)(1 - a) = 1$
Which simplifies to:
$ab - (a+b) = 0$
$ab = a+b$
Meanwhile the area of the triangle is $\frac{1}{2} ab = A$ so $ab = 2A$
Product of roots is 2A
And plugging this back into the earlier equation,
$ab = a+b = 2A $
Sum of roots is also 2A.
So this gives us the final equation $x^2 - 2Ax + 2A = 0$.
A: Let's say the quadratic equation is $ax^2+bx+c=0$.  Without loss of generality, we assume that $a=1$.  Then, the roots are 
$$x_{\pm}=\frac{-b\pm\sqrt{b^2-4c}}{2}$$
The product of the roots are simply 
$$x_+x_-=c$$
and since we must have $\frac12 x_+x_-=A$, then $c=2A$.  Our quadratic equation is now of the form $x^2+bx+2A=0$.  
We also know that the line that passes through $(0,x_+)$ and $(x_-,0)$ also passes through $(1,1)$.  Thus, we have
$$\frac{x_+-1}{1}=\frac{x_+}{x_-}\implies x_+x_-=x_++x_-\implies c=-b\implies b=-c=-2A$$
Thus, the quadratic equation is 
$$\bbox[5px,border:2px solid #C0A000]{x^2-2Ax+2A=0}$$
