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

Question

I do not how to solve this, can such equation even exist? For the root to lie on the y intercept, the line would have to pass through origin, which means one root will be 0, breaking down the whole the thing. Am I missing something here?

Answer 1

Hint:

WLOG equation of the straight line $$\dfrac xa+\dfrac yb=1$$ where $a,b$ are the intercepts

We need to find $a+b, ab$ in terms of $A$

Now $1=\dfrac1a+\dfrac1b\implies a+b=ab$

We also have $$A=\dfrac12|ab|\implies ab=\pm2A$$

Answer 2

You calculate the intersections of the line with the axes, and you get $(a,0)$ and $(0,b)$, such that $|ab|=2A$. Then your quadratic is $(x-a)(x-b)=0$