Consider the equation $y-y_1=m(x-x_1)$. If $m$ and $x_1$ are fixed and different lines are drawn for different values of $y_1$, then

The point $P(a, b)$ lies on the straight line $3x + 2y = 13$ and the point $Q(b, a)$ lies on the straight line $4x - y = 5$. Then the equation of the line $PQ$ is

The equation of the line passing through $(a \cos^3 \theta, a \sin^3 \theta)$ and parallel to the line $x \sec \theta + y \csc \theta = a$ is:

If a line $L$ passing through the point $A(-2, 4)$ makes an angle of $60^{\circ}$ with the positive direction of $X$-axis in the anti-clockwise direction and $B(p, q)$ lying in the $3^{\text{rd}}$ quadrant is a point on $L$ at a distance of $6$ units from the point $A$,then $\sqrt{p^2+q^2-8q} = $

$A$ straight line $L$ at a distance of $4$ units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of $60^o$ with the line $x + y = 0$. Then an equation of the line $L$ is

Find the equation of the line which intersects the $y$-axis at a distance of $2$ units above the origin and makes an angle of $30^{\circ}$ with the positive direction of the $x$-axis.