Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1992 Upvotes
Verified Answer
The correct answer is:
$3 x+4 y-1=0$
$$
\begin{aligned}
& \text { } 6 x^2-x y-12 y^2=0 \\
& \Rightarrow 6 x^2-3 x y+8 x y-12 y^2=0 \\
& \Rightarrow(2 x-3 y)(3 x+4 y)=0 \ldots \text { (i) } \\
& \Rightarrow 15 x^2+14 x y-8 y^2=0 \\
& \Rightarrow 15 x^2+20 x y-6 x y-8 y^2=0 \\
& \Rightarrow(5 x-2 y)(3 x+4 y)=0 \ldots \text { (ii) }
\end{aligned}
$$
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 $3 \mathrm{x}+4 \mathrm{y}=0$
$$
y-1=\frac{-3}{4}(x+1) \Rightarrow 3 x+4 y-=0
$$
\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
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.