If one of the lines of the pair $ax^{2}+2hxy+by^{2}=0$ bisects the angle between the positive direction of the axes,then $a, b,$ and $h$ satisfy the relation:

If the equation given by $hxy + 10x + 6y + 4 = 0$ represents a pair of lines,then $h$ is equal to

Let $PQR$ be a right-angled isosceles triangle,with the right angle at $Q(2, 1)$. If the equation of the line $PR$ is $2x + y = 3$,then the combined equation representing the pair of lines $PQ$ and $QR$ is:

If the sum of the slopes of the lines represented by $x^2 - 2xy \tan \theta - y^2 = 0$ is $4$,then $\theta =$

Assertion $(A)$: The difference of the slopes of the lines represented by $y^2 - 2xy \sec^2 \alpha + (3 + \tan^2 \alpha)(\tan^2 \alpha - 1) x^2 = 0$ is $4$.
Reason $(R)$: The difference of the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is $\frac{2 \sqrt{h^2 - ab}}{|b|}$.

If one of the lines represented by the equation $ax^2 + 2hxy + by^2 = 0$ is $y = mx$,then