The Distance Formula calculates the straight-line distance between any two points in a coordinate plane, expressed as d = √[(x₂ − x₁)² + (y₂ − y₁)²]. It is a fundamental concept in coordinate geometry, covered in NCERT Class 10 (Chapter 7) and extended in Class 11 and 12. JEE Main and NEET aspirants use this formula repeatedly in problems involving circles, lines, and three-dimensional geometry. This article covers the formula derivation, a complete formula sheet, three solved examples at progressive difficulty levels, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET applications.
Key Distance Formula Expressions at a Glance
Quick reference for the most important distance formulas used in CBSE and competitive exams.
- Distance between two points (2D): \( d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \)
- Distance from origin: \( d = \sqrt{x^2 + y^2} \)
- Distance between two points (3D): \( d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2} \)
- Distance from a point to a line: \( d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \)
- Midpoint formula: \( M = \left(\frac{x_1+x_2}{2},\, \frac{y_1+y_2}{2}\right) \)
- Section formula (internal division): \( P = \left(\frac{mx_2+nx_1}{m+n},\, \frac{my_2+ny_1}{m+n}\right) \)
What is The Distance Formula?
The Distance Formula gives the exact length of the straight line segment connecting two points in a coordinate plane. In simple terms, it answers: how far apart are two given points? The formula is derived directly from the Pythagorean theorem. When two points are plotted on a Cartesian plane, a right triangle can be constructed using the horizontal and vertical differences as legs. The hypotenuse of that triangle is the distance between the points.
In the NCERT curriculum, The Distance Formula is introduced in Class 10, Chapter 7 — Coordinate Geometry. It is revisited and extended to three dimensions in Class 11, Chapter 12 — Introduction to Three Dimensional Geometry. The concept also appears in Class 12 Mathematics in the context of vectors and three-dimensional geometry.
The formula applies in many real-world contexts. Navigation systems, computer graphics, robotics, and map applications all rely on distance calculations between coordinate points. Understanding The Distance Formula deeply helps students solve a wide range of CBSE board problems and forms the backbone of coordinate geometry questions in competitive exams.
The Distance Formula — Expression and Variables
The standard two-dimensional Distance Formula between points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is:
\[ d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \]
| Symbol | Quantity | Unit / Type |
|---|---|---|
| \( d \) | Distance between the two points | Same unit as coordinates (e.g., units, cm, m) |
| \( x_1, y_1 \) | Coordinates of the first point A | Real numbers |
| \( x_2, y_2 \) | Coordinates of the second point B | Real numbers |
| \( (x_2 – x_1) \) | Horizontal difference (run) | Real number |
| \( (y_2 – y_1) \) | Vertical difference (rise) | Real number |
Derivation of The Distance Formula
Consider two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) on a Cartesian plane. Draw a horizontal line from A and a vertical line from B. These two lines meet at point \( C(x_2, y_1) \), forming a right angle at C.
Step 1: Find the length of AC (horizontal leg): \( AC = |x_2 – x_1| \)
Step 2: Find the length of BC (vertical leg): \( BC = |y_2 – y_1| \)
Step 3: Apply the Pythagorean theorem to triangle ABC:
\[ AB^2 = AC^2 + BC^2 = (x_2 – x_1)^2 + (y_2 – y_1)^2 \]
Step 4: Take the positive square root (distance is always non-negative):
\[ d = AB = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \]
This derivation extends naturally to three dimensions by adding a third term \( (z_2 – z_1)^2 \) under the square root.
Complete Coordinate Geometry Formula Sheet
| Formula Name | Expression | Variables | Units | NCERT Chapter |
|---|---|---|---|---|
| Distance Formula (2D) | \( d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \) | Two points in a plane | Length units | Class 10, Ch 7 |
| Distance from Origin | \( d = \sqrt{x^2 + y^2} \) | Point (x, y) from O(0,0) | Length units | Class 10, Ch 7 |
| Distance Formula (3D) | \( d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} \) | Two points in 3D space | Length units | Class 11, Ch 12 |
| Midpoint Formula | \( M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \) | Midpoint of segment AB | Coordinate units | Class 10, Ch 7 |
| Section Formula (Internal) | \( P = \left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right) \) | Point dividing AB in ratio m:n | Coordinate units | Class 10, Ch 7 |
| Section Formula (External) | \( P = \left(\frac{mx_2-nx_1}{m-n}, \frac{my_2-ny_1}{m-n}\right) \) | External division in ratio m:n | Coordinate units | Class 10, Ch 7 |
| Distance: Point to Line | \( d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \) | Point \((x_1,y_1)\), line \(ax+by+c=0\) | Length units | Class 11, Ch 10 |
| Distance Between Parallel Lines | \( d = \frac{|c_1 – c_2|}{\sqrt{a^2 + b^2}} \) | Lines \(ax+by+c_1=0\) and \(ax+by+c_2=0\) | Length units | Class 11, Ch 10 |
| Area of Triangle (Coordinate) | \( A = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| \) | Three vertices of triangle | Square units | Class 10, Ch 7 |
| Slope of a Line | \( m = \frac{y_2 – y_1}{x_2 – x_1} \) | Two points on the line | Dimensionless | Class 11, Ch 10 |
The Distance Formula — Solved Examples
Example 1 (Class 9-10 Level)
Problem: Find the distance between the points A(3, 4) and B(7, 1).
Given: \( x_1 = 3,\; y_1 = 4,\; x_2 = 7,\; y_2 = 1 \)
Step 1: Write The Distance Formula: \( d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \)
Step 2: Substitute the values: \( d = \sqrt{(7-3)^2 + (1-4)^2} \)
Step 3: Simplify inside the square root: \( d = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} \)
Step 4: Evaluate: \( d = 5 \) units
Answer
The distance between A(3, 4) and B(7, 1) is 5 units.
Example 2 (Class 11-12 Level)
Problem: Show that the points P(1, 7), Q(4, 2), and R(−4, 6) are the vertices of an isosceles triangle. Find the type of triangle formed.
Given: P(1, 7), Q(4, 2), R(−4, 6)
Step 1: Calculate PQ using The Distance Formula:
\( PQ = \sqrt{(4-1)^2 + (2-7)^2} = \sqrt{9 + 25} = \sqrt{34} \) units
Step 2: Calculate QR:
\( QR = \sqrt{(-4-4)^2 + (6-2)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \) units
Step 3: Calculate PR:
\( PR = \sqrt{(-4-1)^2 + (6-7)^2} = \sqrt{25 + 1} = \sqrt{26} \) units
Step 4: Compare the sides. Since PQ ≠ QR ≠ PR, check if any two sides are equal. None are equal here, but verify: \( PQ^2 = 34,\; PR^2 = 26,\; QR^2 = 80 \). Since \( PQ^2 + PR^2 = 34 + 26 = 60 \neq 80 \), it is a scalene triangle. However, if we check \( PQ^2 + PR^2 \neq QR^2 \), the triangle is scalene with no special property. The key skill demonstrated here is computing all three side lengths systematically using The Distance Formula.
Answer
PQ = \(\sqrt{34}\) units, QR = \(4\sqrt{5}\) units, PR = \(\sqrt{26}\) units. The triangle is scalene (all sides unequal). This problem demonstrates how to classify triangles using The Distance Formula.
Example 3 (JEE/NEET Level)
Problem: A circle has its centre at C(h, 3) and passes through the points A(1, 1) and B(5, 5). Find the value of h and the radius of the circle.
Given: Centre C(h, 3), points A(1, 1) and B(5, 5) lie on the circle.
Step 1: Since both A and B lie on the circle, CA = CB (radii are equal). Apply The Distance Formula to both:
\( CA = \sqrt{(h-1)^2 + (3-1)^2} = \sqrt{(h-1)^2 + 4} \)
\( CB = \sqrt{(h-5)^2 + (3-5)^2} = \sqrt{(h-5)^2 + 4} \)
Step 2: Set CA = CB and square both sides:
\( (h-1)^2 + 4 = (h-5)^2 + 4 \)
Step 3: Expand and simplify:
\( h^2 – 2h + 1 = h^2 – 10h + 25 \)
\( 8h = 24 \implies h = 3 \)
Step 4: Find the radius using CA with h = 3:
\( r = \sqrt{(3-1)^2 + (3-1)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2} \) units
Answer
h = 3 and the radius of the circle = \( 2\sqrt{2} \) units. This JEE-style problem shows how The Distance Formula is used to find unknown coordinates and circle properties.
CBSE Exam Tips 2025-26
- Label your coordinates clearly. Always write \( x_1, y_1, x_2, y_2 \) before substituting. This prevents sign errors, especially with negative coordinates.
- Squaring removes the sign issue. Remember that \( (x_2 – x_1)^2 = (x_1 – x_2)^2 \). The order of subtraction does not affect the final answer. We recommend writing the larger coordinate first to avoid confusion.
- Simplify surds fully. In CBSE 2025-26 board exams, leaving an answer as \( \sqrt{50} \) instead of \( 5\sqrt{2} \) may cost you marks. Always simplify surds completely.
- Check the question type. Some questions ask you to prove that points form a specific shape (square, rhombus, equilateral triangle). Compute all required distances systematically before drawing conclusions.
- Use the 3D formula in Class 11-12 questions. When coordinates include a z-value, immediately switch to \( d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} \). Forgetting the z-term is a common board exam error.
- Collinearity check. Three points A, B, C are collinear if AB + BC = AC. This is a frequent CBSE application of The Distance Formula. Practice this type thoroughly.
Common Mistakes to Avoid
- Mistake 1 — Forgetting the square root: Some students compute \( (x_2-x_1)^2 + (y_2-y_1)^2 \) and report that as the distance. The square root is mandatory. Distance = \( \sqrt{\text{sum of squares}} \), not the sum itself.
- Mistake 2 — Sign errors with negative coordinates: When a coordinate is negative, squaring it correctly is critical. For example, \( (-3 – 2)^2 = (-5)^2 = 25 \), not \( -25 \). Always bracket negative values carefully.
- Mistake 3 — Mixing up x and y coordinates: Ensure \( x_1 \) is paired with \( x_2 \) and \( y_1 \) with \( y_2 \). Swapping an x-value with a y-value gives a completely wrong answer.
- Mistake 4 — Not simplifying the surd: Leaving the answer as \( \sqrt{72} \) instead of \( 6\sqrt{2} \) is an incomplete answer. CBSE expects fully simplified surds.
- Mistake 5 — Using the 2D formula in 3D problems: When points are given as \( (x, y, z) \), students sometimes apply the 2D formula and ignore the z-coordinates. Always count the number of coordinates in each point before choosing the formula.
JEE/NEET Application of The Distance Formula
In our experience, JEE aspirants encounter The Distance Formula in nearly every topic of coordinate geometry. It is rarely tested in isolation. Instead, it appears embedded in multi-step problems involving circles, parabolas, ellipses, and hyperbolas.
Application Pattern 1 — Circle Problems
The equation of a circle \( (x-h)^2 + (y-k)^2 = r^2 \) is itself a direct application of The Distance Formula. Any point \( (x, y) \) on the circle is at distance \( r \) from the centre \( (h, k) \). JEE problems often ask you to find the centre or radius given conditions, requiring you to equate two distance expressions and solve algebraically (as shown in Example 3 above).
Application Pattern 2 — Locus Problems
A locus is the set of all points satisfying a given distance condition. For example: Find the locus of a point equidistant from A(2, 3) and B(6, 1). Setting \( PA = PB \) using The Distance Formula and simplifying gives the equation of the perpendicular bisector of AB. JEE Main frequently tests this concept. Our experts suggest practising at least ten locus problems before the exam.
Application Pattern 3 — Three-Dimensional Geometry (JEE Advanced)
In JEE Advanced, The Distance Formula extends to 3D. Problems involving the distance between a point and a plane, or between two skew lines, build on the 3D distance formula. The shortest distance between two skew lines uses vector methods, but the foundational idea is still the distance between two points. Mastering the 2D formula first makes the 3D extension straightforward. For NEET, The Distance Formula appears in physics-based coordinate problems, particularly in optics and mechanics when positions of objects are given as coordinates.
FAQs on The Distance Formula
For more coordinate geometry formulas, visit our Complete Algebra Formulas hub. You may also find these related articles helpful: Geometric Series Formula, R-Squared Formula, and Percent Error Formula. For the official NCERT syllabus and textbooks, refer to ncert.nic.in.