The three sides of a triangle are given by $(x^2+7xy+2y^2)(y-1)=0$. Then the centroid of that triangle is

If $\alpha x^2+2 \gamma x y+\beta y^2=0$ is the equation of a pair of lines passing through the origin and perpendicular to the pair of lines $b h x^2+a b x y+a h y^2=0$ $(a \neq 0, b \neq 0)$,then $\frac{\alpha \beta}{\gamma^2}=$

The perpendiculars are drawn to lines $L_1$ and $L_2$ from the origin making an angle $\frac{\pi}{4}$ and $\frac{3 \pi}{4}$ respectively with the positive direction of the $X$-axis. If both the lines are at a unit distance from the origin,then their joint equation is

If one of the lines given by the equation $x^{2}+kxy+2y^{2}=0$ is $x+2y=0$,then $k=$

The combined equation of the pair of lines passing through the origin and perpendicular to the pair of lines $2x^2 - xy - y^2 = 0$ is:

If $O(0,0)$,$A(1,2)$,and $B(3,4)$ are the vertices of triangle $OAB$,then the joint equation of the altitude and median drawn from $O$ is