Question

Find the equations to the pair of lines through the origin which are perpendicular to the lines represented by $2x^2-7xy+3y^2=0$

Answer 1

We have $2x^2-7xy+3y^2=0$

$2x^2-6xy-xy+3y^2=0$

$\Rightarrow$ $2x(x-3y)-y(x-3y)=0$

$\Rightarrow$ $(x-3y)(2x-y)=0$

$\Rightarrow$ $x-3y=0or2x-y=0$

Thus the given equation represents the lines $x-3y=0$ and $2x-y=0$

The equations of the lines passing through the origin and perpendicular to the given lines are y - 0 = -3(x - 0) and y - 0 = $-\frac{1}{2}(x-0)[\because (Slopeofx-3y=0)is1/3and(Slopeof2x-y=0)is2]$

$\Rightarrow$ $y+3x=0$ and $2y+x=0$