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

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

If the angle between the pair of lines $x^2+2 \sqrt{2} x y+k y^2=0, k>0$ is $45^{\circ}$,then the area (in square units) of the triangle formed by the pair of bisectors of angles between the given lines and the line $x+2 y+1=0$ 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

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 the equation of the pair of straight lines intersecting at $(a, b)$ and perpendicular to the pair of lines $3x^2 - 4xy + 5y^2 = 0$ is $lx^2 + 2hxy + my^2 - 32x - 26y + c = 0$,then $\frac{a+b+c}{l+h+m} =$