Diagonals AC and BD of square ABCD intersect at P. Coordinates of points B and D are (9, -2) and (1, 6) respectively. 23(a)(i)Find the co-ordinates of point P.
Show Hint
Because the diagonals of a square bisect each other, the midpoint of $BD$ is identical to the midpoint of $AC$.
If you are ever asked to find the coordinates of another vertex when one is missing, this midpoint equivalence is the most direct tool to use.
Step 1: Find P using a vector step from B towards D, instead of the plain midpoint formula.
Since the diagonals of a square bisect each other, point $P$ is exactly halfway along the segment from $B$ to $D$. We can reach $P$ by starting at $B$ and moving half of the vector from $B$ to $D$.
Step 2: Find the vector from B to D.
\[ \vec{BD} = D - B = (1 - 9,\ 6 - (-2)) = (-8,\ 8) \]
Step 3: Take half of this vector, since P is the midpoint.
\[ \frac{1}{2}\vec{BD} = (-4,\ 4) \]
Step 4: Add this half-vector to point B to reach P.
\[ P = B + \frac{1}{2}\vec{BD} = (9 + (-4),\ -2 + 4) = (5,\ 2) \]
Final Answer:
The coordinates of point $P$ are $(5, 2)$.
\[ \boxed{P = (5,2)} \]
