Geometric Applications of the Distance Formula for CBSE Class 10

The distance formula is not merely a tool for calculating lengths between pairs of points; it is a powerful analytical bridge that enables us to identify and verify geometric figures on the Cartesian plane. Given the coordinate vertices of three or four points, we can determine whether they form a specific type of triangle or quadrilateral without making physical measurements.

In CBSE Class 10 Mathematics, Chapter 7 (Coordinate Geometry), questions testing the geometric applications of the distance formula are standard 3-mark and 4-mark board examination problems.

What You Will Learn

• Criteria for proving types of triangles: Equilateral, Isosceles, Right-angled, and Isosceles-right
• Criteria for proving types of quadrilaterals: Parallelogram, Rectangle, Rhombus, and Square
• The decisive role of diagonals in distinguishing rectangles from parallelograms, and squares from rhombuses
• Step-by-step solved CBSE board examination proofs
• Methodical presentation templates and common student traps

1. Classification of Triangles Using Distance

Let $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ be three non-collinear points representing the vertices of $\Delta ABC$. Calculate the lengths of the three sides: $AB$, $BC$, and $CA$.

2. Classification of Quadrilaterals Using Distance

To determine the exact geometric nature of a quadrilateral $ABCD$, calculating only the four outer sides ($AB, BC, CD, DA$) is NOT sufficient! You must ALWAYS calculate the lengths of the two diagonals ($AC$ and $BD$).

                         Classifying Quadrilaterals
                                     |
       +-----------------------------+-----------------------------+
       |                                                           |
Opposite Sides Equal                                       All Four Sides Equal
(AB = CD and BC = DA)                                      (AB = BC = CD = DA)
       |                                                           |
   Diagonals:                                                  Diagonals:
   - If AC = BD  →  RECTANGLE                                  - If AC = BD  →  SQUARE
   - If AC ≠ BD  →  PARALLELOGRAM                              - If AC ≠ BD  →  RHOMBUS

The Decisive Summary Table:

Important: <u>A rectangle is a parallelogram with equal diagonals. A square is a rhombus with equal diagonals. In board examinations, if you prove all four sides are equal but forget to check the diagonals, you will lose marks because you have only proven it is a rhombus, not a square!</u>

3. Solved CBSE Board Examination Problems

Solved Example 1: Proving a Square (NCERT Classic)

Problem: Show that the points $A(1, 7)$, $B(4, 2)$, $C(-1, -1)$, and $D(-4, 4)$ are the vertices of a square.

Solution: We must calculate 6 distances: 4 sides ($AB, BC, CD, DA$) and 2 diagonals ($AC, BD$).

Step 1: Calculate the Four Sides

1. $AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$
2. $BC = \sqrt{(-1 - 4)^2 + (-1 - 2)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$
3. $CD = \sqrt{[-4 - (-1)]^2 + [4 - (-1)]^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$
4. $DA = \sqrt{[1 - (-4)]^2 + (7 - 4)^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34}$

Conclusion 1: All four sides are equal: $AB = BC = CD = DA = \sqrt{34}$.

Step 2: Calculate the Two Diagonals

1. $AC = \sqrt{(-1 - 1)^2 + (-1 - 7)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68}$
2. $BD = \sqrt{(-4 - 4)^2 + (4 - 2)^2} = \sqrt{(-8)^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68}$

Conclusion 2: Both diagonals are equal: $AC = BD = \sqrt{68}$.

Step 3: Final Verification

Since all four sides are equal ($AB = BC = CD = DA$) and both diagonals are equal ($AC = BD$), <u>the quadrilateral $ABCD$ is a square</u>.

Solved Example 2: Proving a Right-Angled Triangle

Problem: Check whether $(5, -2)$, $(6, 4)$, and $(7, -2)$ are the vertices of an isosceles triangle.

Solution: Let $A(5, -2)$, $B(6, 4)$, and $C(7, -2)$.

1. Calculate $AB$: $AB = \sqrt{(6 - 5)^2 + [4 - (-2)]^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37}$
2. Calculate $BC$: $BC = \sqrt{(7 - 6)^2 + (-2 - 4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37}$
3. Calculate $CA$: $CA = \sqrt{(7 - 5)^2 + [-2 - (-2)]^2} = \sqrt{2^2 + 0^2} = \sqrt{4} = 2$
4. Since $AB = BC = \sqrt{37} \ne CA$, exactly two sides are equal.
5. Therefore, <u>$A, B,$ and $C$ are the vertices of an isosceles triangle</u>.

4. Summary and Examination Tips

Exam Tip: In questions asking to prove a rectangle or square, leaving out the calculation of diagonals will automatically result in a $1$-mark penalty because equal sides alone only prove a parallelogram or rhombus!

Common Mistake: Calculating the wrong diagonals! In quadrilateral $ABCD$, vertices must be taken in cyclic order ($A \to B \to C \to D$). The diagonals are always $AC$ and $BD$ (connecting non-consecutive vertices). Never calculate $AB$ or $CD$ as diagonals!