The straight line passing through $(-1,1)$ and remaining parallel to the line common to the pairs of lines…
The straight line passing through $(-1,1)$ and remaining parallel to the line common to the pairs of lines provided by $6 x^2-x y-12 y^2=0$ and $15 x^2+14 x y-8 y^2=0$, is
$5 \mathrm{x}-2 \mathrm{y}+7=0$
$3 x+4 y-1=0$
$3 x-4 y+7=0$
$2 x-3 y+5=0$
Solution
$
\begin{aligned}
& \text { } 6 x^2-x y-12 y^2=0 \\
& \Rightarrow 6 x^2-3 x y+8 x y-12 y^2=0 \\
& \Rightarrow(2 x-3 y)(3 x+4 y)=0 \ldots \text { (i) } \\
& \Rightarrow 15 x^2+14 x y-8 y^2=0 \\
& \Rightarrow 15 x^2+20 x y-6 x y-8 y^2=0 \\
& \Rightarrow(5 x-2 y)(3 x+4 y)=0 \ldots \text { (ii) }
\end{aligned}
$
Equation of common line from (i) and (ii) is
$
\begin{aligned}
& 3 \mathrm{x}+4 \mathrm{y}=0 \\
& \Rightarrow \mathrm{y}=\frac{-3}{4} \mathrm{x} \Rightarrow \operatorname{slope}(\mathrm{m})=\frac{-3}{4}
\end{aligned}
$
Equation of straight line passes through $(-1,1)$ and parallel to $3 \mathrm{x}+4 \mathrm{y}=0$
$
y-1=\frac{-3}{4}(x+1) \Rightarrow 3 x+4 y-=0
$