In quadrilateral PQRS, the coordinates are P(0, 0), Q(a + c, 0), R(2a + c, b), and S(a, b). How can you use coordinate geometry to show that the diagonals are perpendicular?

Answer 1

To show that the diagonals of quadrilateral \(PQRS\) are perpendicular using coordinate geometry, we can follow these steps:

1. **Identify the diagonals:**
The diagonals of the quadrilateral \(PQRS\) are \(PR\) (connecting \(P\) and \(R\)) and \(QS\) (connecting \(Q\) and \(S\)).

2. **Find the slopes of the diagonals:**
The slope of a line segment connecting two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

- **Slope of \(PR\):**
Points \(P(0, 0)\) and \(R(2a + c, b)\):
\[
m_{PR} = \frac{b - 0}{(2a + c) - 0} = \frac{b}{2a + c}
\]

- **Slope of \(QS\):**
Points \(Q(a + c, 0)\) and \(S(a, b)\):
\[
m_{QS} = \frac{b - 0}{a - (a + c)} = \frac{b}{a - a - c} = \frac{b}{-c} = -\frac{b}{c}
\]

3. **Check if the slopes are negative reciprocals:**
Two lines are perpendicular if the product of their slopes is \(-1\). We need to verify if:
\[
m_{PR} \cdot m_{QS} = -1
\]

Substitute the slopes calculated:
\[
\frac{b}{2a + c} \cdot -\frac{b}{c} = -1
\]

Simplify the left-hand side:
\[
-\frac{b^2}{(2a + c)c}
\]

For the product to be \(-1\), we need:
\[
-\frac{b^2}{(2a + c)c} = -1
\]

This implies:
\[
\frac{b^2}{(2a + c)c} = 1
\]

Cross-multiplying gives:
\[
b^2 = (2a + c)c
\]

This is the condition that must be satisfied for the diagonals to be perpendicular.

Thus, if \( b^2 = (2a + c)c \), the diagonals \(PR\) and \(QS\) are perpendicular.

If this condition is satisfied, we conclude that quadrilateral \(PQRS\) has perpendicular diagonals.