The $x$-intercept of a line passing through the points $\left(-\frac{1}{2}, 1\right)$ and $B(1, 3)$ is

In the figure,a time-distance graph of a linear motion is given. Two positions of time and distance are recorded as: when $T = 0, D = 2$ and when $T = 3, D = 8$. Using the concept of slope,find the law of motion,i.e.,how distance depends upon time.

If the line $\frac{x}{a} + \frac{y}{b} = 1$ passes through the points $(2, -3)$ and $(4, -5)$,then $(a, b) = $

$\beta$ is the angle made by the perpendicular drawn from the origin to the line $L \equiv x+y-2=0$ with the positive $X$-axis in the anticlockwise direction. If '$a$' is the $X$-intercept of the line $L=0$ and $p$ is the perpendicular distance from the origin to the line $L=0$,then $a \tan \beta + p^2 =$

The equation of the line passing through the point $(-3, 1)$ and bisecting the angle between the coordinate axes in the second quadrant is:

The equations of the diagonals of the square formed by the lines $x = 0,$ $y = 0,$ $x = 1,$ and $y = 1$ are