The pair of straight lines $x^2 - 4xy + y^2 = 0$ together with the line $x + y + 4 = 0$ form a triangle which 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):

The pairs of straight lines $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$ form a

The line $x+y=1$ meets the lines represented by the equation $y^3-6xy^2+11x^2y-6x^3=0$ at the points $P, Q, R$. If $O$ is the origin,then $(OP)^2+(OQ)^2+(OR)^2=$

The figure formed by the pairs of lines $6x^2+13xy+6y^2=0$ and $6x^2+13xy+6y^2+10x+10y+4=0$ is a

$A$ 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