The line passing through the points $(1, 4)$ and $(-5, 1)$ intersects the line $4x + 3y - 5 = 0$ at the point:

If the normal form of the equation of a straight line $4x + 3y + 2 = 0$ is $x \cos \alpha + y \sin \alpha = p$ and its intercept form is $\frac{x}{a} + \frac{y}{b} = 1$,then $\frac{p \sec \alpha}{ab} = $

$A$ line $L$ perpendicular to the line $5x - 12y + 6 = 0$ makes a positive intercept on the $Y$-axis. If the distance from the origin to the line $L$ is $2$ units and the angle made by the perpendicular drawn from the origin to the line $L$ with the positive $X$-axis is $\theta$,then $\tan \theta + \cot \theta =$

If the slope of a line is $2$ and it makes an intercept of $-4$ on the $y$-axis,then its equation is:

Find the equation of the line passing through the point $(1, -2)$ and making equal intercepts on the axes.

Find the equation of the line which passes through $(2, 2\sqrt{3})$ and is inclined with the $x-$axis at an angle of $75^{\circ}$.