Question

If a pair of perpendicular straight lines is drawn through origin forming an isosceles triangle right angled at the origin with the line $2x+3y=6$, then equation of the pair of lines is

• A. $5y^2+24xy-5x^2=0$
• B. $5y^2-24xy+5x^2=0$
• C. $5y^2+24xy+5x^2=0$
• D. $5y^2-24xy-5x^2=0$

Answer 1

The correct option is A $5y^2+24xy-5x^2=0$

Let, $y=m_{1}x$ and $y=m_{2}x$ are equation of line passing through origin
$\therefore y^2-(m_1+m_2)xy-x^2=0$
Given $y=m_{1}x$ and $y=m_{2}x$ are perpendicular lines.
$\Rightarrow m_1{m}_2=-1$
$\therefore y^2-x^2-(m_1+m_2)xy=0$
and $\frac{2x+3y}{6}=1$
OAB is isosceles right angle triangle
$\therefore tan\theta =\left|\frac{m_1-\left(\frac{-2}{3}\right)}{1-\frac{2}{3}{m}_1}\right|=tan{45}^{\circ }$
$m_1+\frac{2}{3}=1-\frac{2}{3}{m}_1$ or $m_2+\frac{2}{3}=\frac{2}{3}{m}_2-1$
$\frac{5}{3}{m}_1=\frac{1}{3}$ $\frac{4}{3}=\frac{-m_2}{3}$
$m_1=\frac{1}{5}$ $m_2=-5$
$\therefore y^2-x^2-(-5+\frac{1}{5})xy=0$
$\Rightarrow 5y^2-5x^2+24xy=0$