Write the equation of the lines for which $\tan \theta = \frac{1}{2}$,where $\theta$ is the inclination of the line and $x$-intercept is $4$.

$A$ line through $A(-5, -4)$ meets the lines $x + 3y + 2 = 0$,$2x + y + 4 = 0$,and $x - y - 5 = 0$ at $B$,$C$,and $D$ respectively. If $\left( \frac{15}{AB} \right)^2 + \left( \frac{10}{AC} \right)^2 = \left( \frac{6}{AD} \right)^2$,then the equation of the line is

If the three points are $A(2, 3)$,$B(3, 1)$,and $C(5, 3)$,then the slope of the line passing through $A$ and bisecting $BC$ is:

The straight line passing through $(-3, 6)$ and the midpoint of the line segment joining the points $(4, -5)$ and $(-2, 9)$ has an inclination of:

The number of lines that are parallel to $2x + 6y + 7 = 0$ and have an intercept of length $10$ between the coordinate axes is

In 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