Q.Find the equation of the line passing through $(0, -3)$ and having an inclination of $30^\circ$.

Let the $x$- and $y$-intercepts be $a$ and $b$; solve $a+b=1,\ ab=-6$ as a quadratic to find the two possible intercept pairs, then write $x/a+y/b=1$ for each.

Let the line have $x$-intercept $a$ and $y$-intercept $b$. We're given $a+b=1$ and $ab=-6$.

Then $a$ and $b$ are roots of the quadratic equation $t^2-(a+b)t+ab=0$, i.e.

$t^2 - t - 6 = 0$

Factor: $(t-3)(t+2) = 0$, so $t=3$ or $t=-2$.

So $\{a,b\} = \{3,-2\}$, giving two cases.

The equation of a line with $x$-intercept $a$ and $y$-intercept $b$ is

$\frac{x}{a}+\frac{y}{b} = 1$

Case 1: $a=3,\ b=-2$:

$\frac{x}{3}+\frac{y}{-2} = 1 \implies \frac{x}{3}-\frac{y}{2}=1$

Multiply by 6: $2x - 3y = 6$.

Case 2: $a=-2,\ b=3$:

$\frac{x}{-2}+\frac{y}{3} = 1$

Multiply by 6: $-3x+2y=6$, i.e. $3x-2y=-6$.

So there are two lines satisfying the condition: $2x-3y=6$ and $3x-2y=-6$.