Question

The equation of the pair of perpendicular lines passing through origin and forming an isosceles triangle with the line $2x+3y=6$, is

• A. $5x^2-24xy-5y^2=0$
• B. $4x^2-12xy-4y^2=0$
• C. $6x^2-5xy-6y^2=0$
• D. $9x^2+5xy-9y^2=0$

Answer 1

Answer: (A)

Equation of line $BC$ is $2x+3y=6$

Slope of $BC=\frac{-2}{3}$
Slope of $AB=m_1\tan 45^{\circ }=\left|\frac{m_1+\frac{2}{3}}{1-\frac{2}{3}{m}_1}\right|$
$1-\frac{2}{3}{m}_1=m_1+\frac{2}{3}$
$\Rightarrow m_1\left(1+\frac{2}{3}\right)=1-\frac{2}{3}$
$\Rightarrow m_1=\frac{1}{5}$
$\because$ Line $AB$ and $AC$ are perpendicular
$\therefore m_2=-5$
Equation of line $AB$ is $y=\frac{1}{5}x$
$\Rightarrow 5y-x=0$
Equation of line $AC$ is $y=-5x$
$\Rightarrow y+5x=0$
Combining equation is $(5y-x)(y+5x)=0$
$\Rightarrow 5y^2+24xy-5x^2=0$
$\Rightarrow 5x^2-24xy-5y^2=0$