The line $\frac{x}{3}+\frac{y}{2}=1$ and a pair of lines both passing through the origin form an isosceles triangle. If this pair of lines are perpendicular,then the equation of the pair of straight lines is

$A$ 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

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

If the pairs of straight lines represented by $3x^2+2hxy-3y^2=0$ and $3x^2+2hxy-3y^2+2x-4y+c=0$ form a square,then $(h, c) =$

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