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

The area (in sq units) of the quadrilateral formed by two pairs of lines $\lambda^2 x^2 - m^2 y^2 - n(\lambda x + my) = 0$ and $\lambda^2 x^2 - m^2 y^2 + n(\lambda x + my) = 0$ is:

The area bounded by the angle bisectors of the lines ${x^2} - {y^2} + 2y = 1$ and the line $x + y = 3$ is:

From the point $(3,-4)$,perpendicular lines $L_1$ and $L_2$ are drawn to each of the lines represented by $S \equiv 2x^2+3xy-2y^2-7x+y+3=0$. The area of the quadrilateral formed by the pair of lines $S=0$,$L_1$,and $L_2$ is (in square units):

If the equation of the pair of lines passing through $(1, 1)$ and perpendicular to the pair of lines $2x^2 + xy - y^2 - x + 2y - 1 = 0$ is $ax^2 + 2hxy + by^2 + 2gx + 3y = 0$,then $\frac{b}{a} =$