If $m=1$ is the slope of a line $L$,then the product of the slopes of non-parallel lines which are inclined at an angle of $60^{\circ}$ with $L$ is

If $\frac{1}{ab'} + \frac{1}{ba'} = 0,$ then the lines $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{b'} + \frac{y}{a'} = 1$ are

The equation of the lines which pass through the point $(3, -2)$ and are inclined at $60^o$ to the line $\sqrt{3}x + y = 1$ is:

The line passing through the points $(h, 3)$ and $(4, 1)$ intersects the line $7x - 9y - 19 = 0$ at a right angle. Find the value of $h$. (in $/9$)

The equations of the lines passing through $(1,1)$ and making an angle of $45^{\circ}$ with the line $x+y=0$ are

$A$ line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines $L_1: 2x + y + 6 = 0$ and $L_2: 4x + 2y - p = 0, p > 0$,at the points $A$ and $B$,respectively. If $AB = \frac{9}{\sqrt{2}}$ and the foot of the perpendicular from the point $A$ on the line $L_2$ is $M$,then $\frac{AM}{BM}$ is equal to