Let $O=(0,0)$ : let $A$ and $B$ be points respectively on $X$-axis and $Y$-axis such that $\angle O B A=60^{\circ}$. Let $D$ be a point in the first quadrant such that $A D$ is an equilateral triangle. Then, the slope of $D B$ is

$A(-1, 1)$, $B(5, 3)$ are opposite vertices of a square in $xy$-plane. The equation of the other diagonal (not passing through $(A, B)$ of the square is given by

Let $PS$ be the median of the triangle with vertices $P(2,\;2),\;Q(6,\; - \;1)$ and $R(7,\;3)$. The equation of the line passing through $(1, -1)$ and parallel to $PS$ is

If the line $3x + 3y -24 = 0$ intersects the $x-$ axis at the point $A$ and the $y-$ axis at the point $B$, then the incentre of the triangle $OAB$, where $O$ is the origin, is

The origin and the points where the line $L_1$ intersect the $x$ -axis and $y$ -axis are vertices of right angled triangle $T$ whose area is $8$. Also the line $L_1$ is perpendicular to line $L_2$ : $4x -y = 3$, then perimeter of triangle $T$ is -

The triangle formed by ${x^2} - 9{y^2} = 0$ and $x = 4$ is