Two sides of a rhombus are along the lines, $x -y+ 1 = 0$ and $7x-y-5 =0.$ If its diagonals intersect at $(-1,-2),$ then which one of the following is a vertex of this rhombus?

One side of a rectangle lies along the line $4x + 7y + 5 = 0.$ Two of its vertices are $(-3, 1)$ and $(1, 1)$. Then the equations of other three sides are

Let two points be $\mathrm{A}(1,-1)$ and $\mathrm{B}(0,2) .$ If a point $\mathrm{P}\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)$ be such that the area of $\Delta \mathrm{PAB}=5\; \mathrm{sq}$ units and it lies on the line, $3 x+y-4 \lambda=0$ then a value of $\lambda$ is

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

The equation to the sides of a triangle are $x - 3y = 0$, $4x + 3y = 5$ and $3x + y = 0$. The line $3x - 4y = 0$ passes through

Let $m, n$ be real numbers such that $0 \leq m \leq \sqrt{3}$ and $-\sqrt{3} \leq n \leq 0$. The minimum possible area of the region of the plane consisting of points $(x, y)$ satisfying in inequalities $y \geq 0, y-3 \leq m x$, $y -3 \leq nx$, is