The lines represented by the equations $23x^2 - 48xy + 3y^2 = 0$ and $2x + 3y + 4 = 0$ form

The area (in square units) of the triangle formed by the line $x+y+1=0$ and the pair of straight lines $x^2-3xy+2y^2=0$ is

The distance from the origin to the orthocentre of the triangle formed by the lines $x+y-1=0$ and $6x^2-13xy+5y^2=0$ is

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

Suppose that the sides passing through the vertex $(\alpha, \beta)$ of a triangle are bisected at right angles by the lines $y^2-8xy-9x^2=0$. Then,the centroid of the triangle is

If $P$ is the set of all real numbers $\alpha$ such that the product of the lengths of perpendiculars from $(\alpha, 1)$ to the pair of straight lines $3x^2+7xy+2y^2=0$ is $\frac{\sqrt{2}}{5}$,then the sum of the elements in $P$ is