The equations $x-y=4$ and $x^2+4xy+y^2=0$ represent the sides of a/an

The equation of a line which makes an angle of $45^{\circ}$ with each of the pair of lines $xy-x-y+1=0$ is

Suppose the pairs of straight lines $2x^2 + axy + 3y^2 = 0$ and $2x^2 + bxy - 3y^2 = 0$ are such that they have one common line,and the other two lines are perpendicular. Then the values of $a$ and $b$ are respectively:

$A$ 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

If $ax^2+6xy-2y^2=0$ represents a pair of perpendicular lines and $9x^2+2hxy+4y^2=0$ $(h>0)$ represents a pair of coincident lines,then $h=$