a,b,c Of a Straight line

1. Take the general equation: ax + by + c = 0

2. Write it as a function of x:

   ${y = (- \dfrac{a}{b}}){x}\ + (- \dfrac{c}{b})$

3. The question indicates that the line's slope is positive, so:

${- \dfrac{a}{b}} > {0}$ and $\boxed{{\dfrac{a}{b} < 0}} (1)$ 4. The question also says that the y-intercept is positive, so:

${- \dfrac{c}{b}} > {0}$ and $\boxed{{\dfrac{c}{b} < 0}} (2)$ 5. As (1) and (2) fractions are negative , we can multiply (1) and (2) and get a positive number:

${\dfrac{a}{b} \times \dfrac{c}{b}} > {0}$ and $\boxed{{\dfrac{ac}{b^2}} > {0}}$ 6. By definition , $b^2\ > 0$ for any b, so we must have ac > 0

$\large{\boxed{ac >0}}$