Diagonals AC and BD of square ABCD intersect at P. Coordinates of points B and D are (9, -2) and (1, 6) respectively. (i) Find the coordinates of point P. (ii) Find the length of the side of the square.
Show Hint
Always remember that diagonals of a square are equal and perpendicular bisectors of each other.
Using the relation \( d = s\sqrt{2} \) directly gives the side length \( s \) without needing to use trigonometric functions or resolving the coordinates of vertices \( A \) and \( C \).
Step 1: Find P, the midpoint of diagonal BD.
In a square, the two diagonals bisect each other, so the point P where AC and BD cross is the midpoint of BD (it is also the midpoint of AC).
With $B(9, -2)$ and $D(1, 6)$:
\[ P = \left(\frac{9+1}{2}, \frac{-2+6}{2}\right) = \left(\frac{10}{2}, \frac{4}{2}\right) = (5, 2) \]
Step 2: Find the half diagonal length PB directly, without first finding the whole diagonal BD.
\[ PB = \sqrt{(9-5)^2 + (-2-2)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \text{ units} \]
Step 3: Use the right angle at P to build a right triangle with the side of the square.
The diagonals of a square cross at right angles and are equal in length, so $PB = PC$ (both are half diagonals). At vertex P, the segments PB and PC form a right angled triangle PBC in which the side BC of the square is the hypotenuse:
\[ BC^2 = PB^2 + PC^2 = PB^2 + PB^2 = 2 \times PB^2 \]
Step 4: Substitute PB and solve for the side of the square.
\[ BC^2 = 2 \times (4\sqrt{2})^2 = 2 \times 32 = 64 \]
\[ BC = \sqrt{64} = 8 \text{ units} \]
Final Answer:
(i) The coordinates of point P are $(5, 2)$.
(ii) The side of the square is
\[ \boxed{8 \text{ units}} \]
