Reduce the equation $y-2=0$ into the normal form $x \cos \omega + y \sin \omega = p$. Find the perpendicular distance from the origin $(p)$ and the angle between the perpendicular and the positive $x$-axis $(\omega)$.

$A$ straight line cuts off the intercepts $OA = a$ and $OB = b$ on the positive directions of $x$-axis and $y$-axis respectively. If the perpendicular from origin $O$ to this line makes an angle of $\frac{\pi}{6}$ with the positive direction of $y$-axis and the area of $\triangle OAB$ is $\frac{98}{3} \sqrt{3}$,then $a^2 - b^2$ is equal to:

The equation of the line,which bisects the line segment joining the points $(2, -19)$ and $(6, 1)$ and is perpendicular to the line joining the points $(-1, 3)$ and $(5, -1)$,is

If a line intercepted between the coordinate axes is trisected at a point $A(4, 3)$,which is nearer to the $x$-axis,then its equation is

The gradient of the line joining the points on the curve $y = x^2 + 2x$ whose abscissae are $1$ and $3$,is

The equations of the lines which cut off an intercept of $-1$ from the $y$-axis and are equally inclined to the axes are