If two pairs of straight lines with combined equations $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$ form a square,then the combined equation of its diagonals is

The centroid of the triangle formed by the pair of straight lines $12x^2 - 20xy + 7y^2 = 0$ and the line $2x - 3y + 4 = 0$ is:

If a pair of lines drawn through the origin forms an isosceles right-angled triangle with the line $2x + 3y = 6$,then those lines are

Let $P$ be the pair of lines represented by $2x^2 - 5xy + 2y^2 + 6x - 3y = 0$. Consider the following independent statements:
$(i)$ $\alpha$ is the $x$-coordinate of the point of intersection of the pair of lines $P$.
(ii) $\beta$ is the slope of one of the lines of $P$ passing through the origin.
(iii) $\gamma$ is the constant term in the equation of the pair of angular bisectors of $P$.
Then,

The numbers $\alpha$ and $\beta$ are such that one of the lines of $2x^2 + \alpha xy + 3y^2 = 0$ coincides with one of the lines of $2x^2 + \beta xy - 3y^2 = 0$. If the two lines other than that common line are perpendicular,then $|\alpha + \beta|$ is equal to

Area of the triangle formed by the lines $3x^2-4xy+y^2=0$ and $2x-y=6$ is