The equation of the line,where the length of the perpendicular segment from the origin to the line is $4$ and the inclination of this perpendicular segment with the positive direction of the $X$-axis is $30^{\circ}$,is:

If the straight line $2x - 3y + 17 = 0$ is perpendicular to the line passing through the points $(7, 17)$ and $(15, \beta)$,then $\beta$ equals

Reduce the following equation into intercept form and find its intercepts on the axes: $3x + 2y - 12 = 0$.

The equation of the line passing through the origin and the point $(a \cos \theta, a \sin \theta)$ is .....

$A$ 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:

$A$ line passes through $(-1, -3)$ and is perpendicular to $x + 6y = 5$. Its $x$-intercept is