Question

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.

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.

Answer 1

Step 1: Understanding the Question:
This question is from "Coordinate Geometry", utilizing properties of squares.
We are given a square $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $P$.
The coordinates of two opposite vertices, $B(9, -2)$ and $D(1, 6)$, representing the endpoints of diagonal $BD$, are provided.
We need to find the coordinates of the intersection point $P$.

Step 2: Key Formula or Approach:
A fundamental geometric property of any square (and indeed any parallelogram) is that its diagonals bisect each other.
This means the point of intersection $P$ is the midpoint of both diagonals $AC$ and $BD$.
Thus, we can find the coordinates of $P$ using the Midpoint Formula on diagonal $BD$:
\[ P(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

Step 3: Detailed Explanation:

• Identify the coordinates of the endpoints of diagonal $BD$:
- $B(x_1, y_1) = (9, -2)$
- $D(x_2, y_2) = (1, 6)$

• Apply the Midpoint Formula to calculate the x-coordinate of $P$:
\[ x_P = \frac{x_1 + x_2}{2} = \frac{9 + 1}{2} = \frac{10}{2} = 5 \]

• Apply the Midpoint Formula to calculate the y-coordinate of $P$:
\[ y_P = \frac{y_1 + y_2}{2} = \frac{-2 + 6}{2} = \frac{4}{2} = 2 \]

• Combine these components to state the coordinates of the intersection point $P$:
\[ P = (5, 2) \]

Step 4: Final Answer:
The coordinates of the point of intersection $P$ are $(5, 2)$.