Answer
$y=-\dfrac{1}{3}x+1$
$y=-\dfrac{1}{12}x-\dfrac{1}{2}$
Work Step by Step
Let $y=mx+b$ the equation of a line passing through the point $(6,-1)$ and having the product of the $x$- and $y$-intercepts equal to 3.
Because the line passes through $(6,-1)$ we have:
$6m+b=-1$
Because the product of the $x$- and $y$-intercepts is 3, we have:
$b\cdot \left(-\dfrac{b}{m}\right)=3\Rightarrow b^2=-3m$
Use the two equations:
$\begin{cases}
6m+b=-1\\
b^2=-3m
\end{cases}$
$(-1-6m)^2=-3m$
$1+12m+36m^2+3m=0$
$36m^2+15m+1=0$
$m=\dfrac{-15\pm\sqrt{15^2-4(36)(1)}}{2(36)}=\dfrac{-15\pm 9}{72}$
$m_1=\dfrac{-15-9}{72}=-\dfrac{1}{3}$
$m_2=\dfrac{-15+9}{72}=-\dfrac{1}{12}$
Determine $b$:
$m_1=-\dfrac{1}{3}\Rightarrow b=-1-6\left(-\dfrac{1}{3}\right)=-1+2=1$
$m_2=-\dfrac{1}{12}\Rightarrow b=-1-6\left(-\dfrac{1}{12}\right)=-1+\dfrac{1}{2}=-\dfrac{1}{2}$
The lines are:
$y=-\dfrac{1}{3}x+1$
$y=-\dfrac{1}{12}x-\dfrac{1}{2}$
