If $P(4, -7)$ and $Q(-1, 5)$,then $PQ = \dots$

If the vertices of $\Delta ABC$ are $A(0, 0)$,$B(4, 0)$ and $C(0, 6)$,then the coordinates of the point $P$ which is equidistant from $A, B$ and $C$ and in the plane of $\Delta ABC$ is $\ldots \ldots \ldots$

$P$ is a point on $\overline{AB}$ such that $A - P - B$. $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ are the given points. If point $P$ divides $\overline{AB}$ from $A$ in the ratio $m : n$ (where $\frac{m}{n} > 0$),then the coordinates of $P$ are:

The line segment joining the points $A(3, 2)$ and $B(5, 1)$ is divided at the point $P$ in the ratio $1: 2$,and it lies on the line $3x - 18y + k = 0$. Find the value of $k$.

Using the distance formula,show that the points $A(7,3)$,$B(3,0)$,$C(0,-4)$,and $D(4,-1)$ are the vertices of a rhombus.

If $d[(5,0), (x, 4)] = \sqrt{17}$,then $x = \ldots \ldots \ldots$