The lines $(lx + my)^2 - 3(mx - ly)^2 = 0$ and $lx + my + n = 0$ form

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

The difference of the tangents of the angles which the lines $(\tan ^2 \alpha+\cos ^2 \alpha) x^2-2 x y \tan \alpha +(\sin ^2 \alpha) y^2=0$ make with the $X$-axis is

One line of the pair of lines $x^2+xy-2y^2=0$ is perpendicular to one line of the pair of lines $3y^2-5xy-2x^2=0$. If the combined equation of the two lines other than those two perpendicular lines is $ax^2+2hxy+by^2=0$,then $a+2h+b=$

The product of the lengths of the perpendiculars drawn from the point $(-1, 5)$ to the pair of lines $2x^2 - xy - 3y^2 + 6x + y + 4 = 0$ is

If the angle between the pair of lines $x^2+2 \sqrt{2} x y+k y^2=0, k>0$ is $45^{\circ}$,then the area (in square units) of the triangle formed by the pair of bisectors of angles between the given lines and the line $x+2 y+1=0$ is