Question

TOAD is a quadrilateral with vertices (9, 10), (18, 10), (14, 5), and (5, 5), respectively. The diagonals intersect at Point S.

Answer 1

⇒Plotting the Quadrilateral on two Dimensional Plane

 And finding the diagonal of Quadrilateral bisect each other.

If you will look at the Quadrilateral, none of the interior angle is of 90°, so it can't be Square or Rectangle.

Finding the length of two Adjacent Sides

$TO=\sqrt{(18-9)^2+(10-10)^2}\\\\TO=9 \text{Unit}\\\\OA=\sqrt{(18-14)^2+(10-5)^2}\\\\OA=\sqrt{4^2+5^2}\\\\OA=\sqrt{41}$

As, the Length of Adjacent sides are not equal,so it can't be a Rhombus.

The Given Quadrilateral ,"TOAD" is a Parallelogram.

⇒The Point of Intersection of both Diagonal can be Obtained by

       $=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})\\\\\text{Point P}=(\frac{14+9}{2},\frac{5+10}{2})\\\\\text{Point P}=(\frac{23}{2},\frac{15}{2})\\\\\text{Point P}=(\frac{18+5}{2},\frac{10+5}{2})\\\\\text{Point P}=(\frac{23}{2},\frac{15}{2})$

Intersection of both diagonals is same, diagonals bisect each other.    

Point of Intersection of Diagonal is Point P

     $=(\frac{23}{2},\frac{15}{2})$

Part C      

Distance formula

   $=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2)}\\\\TP=\sqrt{(\frac{23}{2}-9)^2+(\frac{15}{2}-10)^2}\\\\TP=\sqrt{\frac{25}{4}+\frac{25}{4}}\\\\TP=\frac{5}{\sqrt{2}}\\\\PA=\sqrt{(\frac{23}{2}-14)^2+(\frac{15}{2}-5)^2}\\\\PA=\sqrt{\frac{25}{4}+\frac{25}{4}}\\\\TP=\frac{5}{\sqrt{2}}\\\\PD=\sqrt{(\frac{23}{2}-5)^2+(\frac{15}{2}-5)^2}\\\\PD=\sqrt{\frac{169}{4}+\frac{25}{4}}\\\\PD=\frac{\sqrt{169+25}}{2}\\\\PD=\frac{\sqrt{194}}{2}\\\\PO=\sqrt{(\frac{23}{2}-18)^2+(\frac{15}{2}-10)^2}\\\\PO=\sqrt{\frac{169}{4}+\frac{25}{4}}\\\\PO=\frac{\sqrt{169+25}}{2}\\\\PO=\frac{\sqrt{194}}{2}$

→DP=PO

→PT=PA

Length of Segments from point of intersection of two diagonal is same.So,  Diagonal are bisector of each other.

Answer 2

A. parallelogram
B. I believe it'd be around (11.2,7.5)
C. √(10-5)^2+(9-4)^2 = 21, so yes they bisect