Search any question & find its solution
Question:
Answered & Verified by Expert
The equation of the pair of perpendicular lines passing through origin and forming an isosceles triangle with the line $2 x+3 y=6$, is
Options:
Solution:
1466 Upvotes
Verified Answer
The correct answer is:
$5 x^2-24 x y-5 y^2=0$
Equation of line $B C$ is $2 x+3 y=6$.
Slope of $B C=\frac{-2}{3}$
$$
\begin{aligned}
& \text { Slope of } A B=m_1 \tan 45^{\circ}=\left|\frac{m_1+\frac{2}{3}}{1-\frac{2}{3} m_1}\right| \\
& \quad 1-\frac{2}{3} m_1=m_1+\frac{2}{3} \Rightarrow m_1\left(1+\frac{2}{3}\right)=1-\frac{2}{3} \\
& \Rightarrow m_1=\frac{1}{5}
\end{aligned}
$$
$\because$ Line $A B$ and $A C$ are perpendicular
$$
\therefore m_2=-5
$$
Equation of line $A B$ is $y=\frac{1}{5} x \Rightarrow 5 y-x=0$
Equation of line $A C$ is $y=-5 x \Rightarrow y+5 x=0$
Combining equation is $(5 y-x)(y+5 x)=0$
$$
\Rightarrow 5 y^2+24 x y-5 x^2=0 \Rightarrow 5 x^2-24 x y-5 y^2=0
$$
Slope of $B C=\frac{-2}{3}$
$$
\begin{aligned}
& \text { Slope of } A B=m_1 \tan 45^{\circ}=\left|\frac{m_1+\frac{2}{3}}{1-\frac{2}{3} m_1}\right| \\
& \quad 1-\frac{2}{3} m_1=m_1+\frac{2}{3} \Rightarrow m_1\left(1+\frac{2}{3}\right)=1-\frac{2}{3} \\
& \Rightarrow m_1=\frac{1}{5}
\end{aligned}
$$
$\because$ Line $A B$ and $A C$ are perpendicular
$$
\therefore m_2=-5
$$
Equation of line $A B$ is $y=\frac{1}{5} x \Rightarrow 5 y-x=0$
Equation of line $A C$ is $y=-5 x \Rightarrow y+5 x=0$
Combining equation is $(5 y-x)(y+5 x)=0$
$$
\Rightarrow 5 y^2+24 x y-5 x^2=0 \Rightarrow 5 x^2-24 x y-5 y^2=0
$$
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.