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

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

The orthocenter of the triangle formed by the lines $x-2y=10$ and $6x^2+xy-y^2=0$ is

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

If the equation of the pair of lines passing through $(1, 1)$ and perpendicular to the pair of lines $2x^2 + xy - y^2 - x + 2y - 1 = 0$ is $ax^2 + 2hxy + by^2 + 2gx + 3y = 0$,then $\frac{b}{a} =$

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: