$A$ line has slope $m$ and $y$-intercept $4$. The distance between the origin and the line is equal to

The equations of the lines passing through the point of intersection of the lines $x - y + 1 = 0$ and $2x - 3y + 5 = 0$ and whose distance from the point $(3, 2)$ is $\frac{7}{5}$ are:

$A$ straight line makes an intercept on the $Y$-axis twice as long as that on the $X$-axis and is at a unit distance from the origin. Then the line is represented by the equations:

The angle made by a line $L$ with the positive $X$-axis measured in the positive direction is $\frac{\pi}{6}$ and the intercept made by $L$ on the $Y$-axis is negative. If $L$ is at a distance of $5$ units from the origin,then the perpendicular distance from the point $(1, -\sqrt{3})$ to the line $L$ is

Find the two points on the line $x + y = 4$ which are at a unit distance from the line $4x + 3y = 10$.

If $p$ and $q$ are the lengths of the perpendiculars from the origin to the lines $x \cos \theta - y \sin \theta = k \cos 2 \theta$ and $x \sec \theta + y \csc \theta = k$ respectively,prove that $p^{2} + 4q^{2} = k^{2}$.