Question

The vertices of a quadrilateral ABCD are A(6, – 2), B(9, 2), C(5, – 1) and D(2, – 5). Prove that ABCD is a rhombus, and not a square.

Answer 1

Step 1: Understanding the properties of a rhombus and square:
- A rhombus is a quadrilateral where all four sides have equal length.
- A square is a special type of rhombus where all four angles are right angles (90°).
To prove ABCD is a rhombus, we need to show that all four sides are of equal length.
To prove it is not a square, we need to show that not all angles are 90°.

Step 2: Calculate the length of all sides:
We use the distance formula to find the length between two points (x₁, y₁) and (x₂, y₂), which is given by:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] For each side of the quadrilateral ABCD:

Side AB:
\[ AB = \sqrt{(9 - 6)^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Side BC:
\[ BC = \sqrt{(5 - 9)^2 + (-1 - 2)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Side CD:
\[ CD = \sqrt{(2 - 5)^2 + (-5 - (-1))^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Side DA:
\[ DA = \sqrt{(6 - 2)^2 + (-2 - (-5))^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Since all four sides have equal length (5 units), ABCD is a rhombus.

Step 3: Proving that ABCD is not a square:
To prove ABCD is not a square, we check if the angles are 90°. We can use the dot product of vectors to check if two sides are perpendicular.
- The vector representing side AB is \( \overrightarrow{AB} = (9 - 6, 2 - (-2)) = (3, 4) \).
- The vector representing side BC is \( \overrightarrow{BC} = (5 - 9, -1 - 2) = (-4, -3) \).

The dot product of vectors AB and BC is:
\[ \overrightarrow{AB} \cdot \overrightarrow{BC} = (3)(-4) + (4)(-3) = -12 - 12 = -24 \]

Since the dot product is not zero, the angle between AB and BC is not 90°. Therefore, ABCD is not a square.

Conclusion:
Since ABCD has all sides equal in length but does not have a right angle, it is a rhombus but not a square.