The orthocentre of the triangle formed by the lines $x+3y=10$ and $6x^2+xy-y^2=0$ is

If $2x^2-5xy+2y^2=0$ represents two sides of a triangle whose centroid is $(1,1)$,then the equation of the third side is

The triangle formed by the lines $2x^2 - 3xy - 2y^2 = 0$ and $3x - y = 7$ is...

If one of the pair of lines $4x^2+6xy+ky^2=0$ is perpendicular to one of the lines represented by $3x^2-5xy+2y^2=0$,then twice the absolute difference of such possible values of $k$ is

$A$ pair of straight lines passes through the point $(1,1)$. One of the lines makes an angle $\theta$ with the positive direction of the $X$-axis and the other makes the same angle with the positive direction of the $Y$-axis. If the equation of the pair of straight lines is $x^2-(a+2)xy+y^2+a(x+y-1)=0$,$a \neq -2$,then the value of $\theta$ is

If $d$ is the distance between the point of intersection of the lines $x^2+4xy+ky^2-4x-10y+3=0$ and the origin,and $p$ is the product of the perpendicular distances from the origin to these lines,then $d^2-20p^2=$