Question

Find the joint equation of the pair of lines passing through the origin, which are perpendicular to the lines represented by $5x^2+2xy-3y^2=0$.

Answer 1

Given homogeneous equation is $5x^2+2xy-3y^2=0$ which is factorisable as

$5x^2+5xy-3xy-3y^2=0$

$\Rightarrow 5x(x+y)-3y(x+y)=0$

$\Rightarrow (x+y)(5x-3y)=0$

$x+y=0$ and $5x-3y=0$ are the two lines represented by the given equation.

$\Rightarrow$Their slopes are $–1$ and $\frac{5}{3}$.

Required two lines are respectively perpendicular to these lines.

$\therefore$ Slopes of required lines are $1$ and $-\frac{3}{5}$ and the lines pass through origin.

$\therefore$ Their individual equations are $y=1\cdot x$ and $y=-\frac{3}{5}x$

$i.e.,x-y=0,3x+5y=0$

$\therefore$ Their joint equation is $(x-y)(3x+5y)=0$

$\Rightarrow 3x^2-3xy+5xy-5y^2=0$

$\Rightarrow 3x^2+2xy-5y^2=0$

Hence $3x^2+2xy-5y^2=0$ is the required joint equation.