Classify Quadrilaterals with the Distance Formula | Class 10

Quick answer: To classify a quadrilateral ABCD from its vertices, find all four sides and both diagonals with the distance formula. All sides equal and diagonals equal → square; all sides equal but diagonals unequal → rhombus; opposite sides equal and diagonals equal → rectangle; opposite sides equal but diagonals unequal → parallelogram (not a rectangle).

The distance formula

The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

For a quadrilateral ABCD (vertices taken in order), calculate six distances:

• Sides: AB, BC, CD, DA
• Diagonals: AC, BD

   D --------- C
   | \       / |
   |   \   /   |
   |     X     |
   |   /   \   |
   | /       \ |
   A --------- B
 diagonals: AC and BD

Decision table

Why the diagonals matter: Four equal sides only tell you the figure is a rhombus. A rhombus is a square only when its diagonals are equal. In the same way, a parallelogram is a rectangle only when its diagonals are equal.

Solved example 1: show that the points form a square (NCERT)

Show that A(1, 7), B(4, 2), C(−1, −1) and D(−4, 4) are the vertices of a square.

Step 1: The four sides.

$AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} = \sqrt{9 + 25} = \sqrt{34}$

$BC = \sqrt{(-1 - 4)^2 + (-1 - 2)^2} = \sqrt{25 + 9} = \sqrt{34}$

$CD = \sqrt{(-4 + 1)^2 + (4 + 1)^2} = \sqrt{9 + 25} = \sqrt{34}$

$DA = \sqrt{(1 + 4)^2 + (7 - 4)^2} = \sqrt{25 + 9} = \sqrt{34}$

All four sides are equal, so ABCD is at least a rhombus.

Step 2: The two diagonals.

$AC = \sqrt{(-1 - 1)^2 + (-1 - 7)^2} = \sqrt{4 + 64} = \sqrt{68}$

$BD = \sqrt{(-4 - 4)^2 + (4 - 2)^2} = \sqrt{64 + 4} = \sqrt{68}$

The diagonals are equal.

Step 3: Conclusion. All four sides are equal ($\sqrt{34}$) and the diagonals are equal ($\sqrt{68}$), so ABCD is a square.

Check: In a square, diagonal² = 2 × side². Here 68 = 2 × 34. ✓

Solved example 2: a parallelogram but not a rectangle

Show that A(−2, −1), B(1, 0), C(4, 3) and D(1, 2) are the vertices of a parallelogram that is not a rectangle.

Step 1: The four sides.

$AB = \sqrt{(1 + 2)^2 + (0 + 1)^2} = \sqrt{10}$

$BC = \sqrt{(4 - 1)^2 + (3 - 0)^2} = \sqrt{18}$

$CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{10}$

$DA = \sqrt{(-2 - 1)^2 + (-1 - 2)^2} = \sqrt{18}$

Opposite sides are equal (AB = CD and BC = DA), so ABCD is a parallelogram.

Step 2: The two diagonals.

$AC = \sqrt{(4 + 2)^2 + (3 + 1)^2} = \sqrt{52}$

$BD = \sqrt{(1 - 1)^2 + (2 - 0)^2} = 2$

Step 3: Conclusion. The diagonals are not equal ($\sqrt{52} \ne 2$), so ABCD is a parallelogram but not a rectangle.

Another check: The mid-point of AC is (1, 1) and the mid-point of BD is also (1, 1). The diagonals bisect each other, which confirms a parallelogram.

Summary table

Exam tips

• Always calculate both diagonals. Four equal sides alone cannot prove a square.
• Take the vertices in order A, B, C, D. The diagonals join opposite vertices: AC and BD.
• Leave answers as surds, such as √34; you do not need decimals to compare them.
• You can compare squared distances (34, 68, …) to avoid writing square roots at every step.

Common mistakes

• Sign errors with negative coordinates, for example writing 1 − (−4) as −3 instead of 5.
• Calling a figure a square after finding only four equal sides.
• Treating AB and CD as diagonals. They are opposite sides.
• Forgetting to state the conclusion in words, such as "hence ABCD is a square".

Frequently Asked Questions

How do you prove that four points form a square?

Use the distance formula to show that all four sides are equal and that both diagonals are equal.

What is the difference between a square and a rhombus in coordinate geometry?

Both have four equal sides. A square also has equal diagonals; a rhombus that is not a square has unequal diagonals.

How do you show a quadrilateral is a parallelogram but not a rectangle?

Show that opposite sides are equal (or that the diagonals bisect each other), and then show that the diagonals are not equal.

Are A(1, 7), B(4, 2), C(−1, −1) and D(−4, 4) the vertices of a square?

Yes. All four sides equal √34 and both diagonals equal √68, so ABCD is a square.

Why must we find the diagonals?

Sides alone cannot tell a square from a rhombus, or a rectangle from a parallelogram. Equal diagonals are what separate them.