Distance Formula

Class 10Coordinate Geometry

The Distance Formula is one of the most important and widely used results in Coordinate Geometry, forming a central topic in Chapter 7 of the CBSE Class 10 Mathematics syllabus. It provides a method to calculate the exact distance between any two points in the Cartesian plane when their coordinates are known. The formula is a direct application of the Pythagoras Theorem to the coordinate system and represents a beautiful bridge between algebra and geometry. Before the development of coordinate geometry by Rene Descartes in the 17th century, calculating distances required physical measurement or complex geometric constructions. The Distance Formula transformed this by allowing distances to be computed purely through arithmetic, using the coordinates of the endpoints. In Class 10, students learn to use the Distance Formula to find the distance between two points, prove geometric properties of figures (such as showing that a quadrilateral is a rhombus, rectangle, or square), check collinearity of points, and solve problems involving the positions of points relative to each other. The formula is also the foundation for the equation of a circle (the set of all points equidistant from a centre point) and is used extensively in higher mathematics, physics, engineering, computer science, and navigation. This comprehensive guide covers the derivation, various forms, applications, and a rich set of solved examples to ensure thorough mastery of this essential topic.

What is Distance Formula - Derivation, Applications, Collinearity & Solved Examples?

Distance Formula: The distance between two points P(x_1, y_1) and Q(x_2, y_2) in the Cartesian plane is given by:

PQ = sqrt[(x_2 - x_1)^2 + (y_2 - y_1)^2]

This formula calculates the length of the straight line segment joining the two points. It works for any two points in the plane, regardless of their position (same quadrant, different quadrants, on the axes, or at the origin).

Special Cases:

CasePointsSimplified Formula
Distance from originO(0,0) and P(x,y)OP = sqrt(x^2 + y^2)
Points on x-axis(x_1, 0) and (x_2, 0)|x_2 - x_1|
Points on y-axis(0, y_1) and (0, y_2)|y_2 - y_1|
Same x-coordinate(a, y_1) and (a, y_2)|y_2 - y_1| (vertical distance)
Same y-coordinate(x_1, b) and (x_2, b)|x_2 - x_1| (horizontal distance)

Important Properties:

  • Distance is always non-negative: PQ >= 0.
  • Distance is zero if and only if the two points coincide: PQ = 0 implies P = Q.
  • Distance is symmetric: PQ = QP (swapping x_1, y_1 with x_2, y_2 gives the same result since we square the differences).
  • The order of subtraction does not matter: (x_2 - x_1)^2 = (x_1 - x_2)^2.

Distance Formula Formula

Distance Formula:

d = sqrt[(x_2 - x_1)^2 + (y_2 - y_1)^2]

Distance from Origin:

d = sqrt(x^2 + y^2) for point (x, y) from origin (0, 0)

Applications of Distance Formula:

ApplicationMethod
Length of a line segmentApply the formula directly to the endpoints
Checking collinearityIf AB + BC = AC, then A, B, C are collinear
Type of triangleCompute all three sides; check for equal sides, right angle
Type of quadrilateralCompute all four sides and both diagonals
Equation of a circleSet distance from centre to point = radius
Finding a point given distanceSet up equation using the formula and solve

Identifying Quadrilaterals:

QuadrilateralConditions (sides and diagonals)
ParallelogramOpposite sides equal: AB = CD, BC = DA
RectangleOpposite sides equal AND diagonals equal
RhombusAll four sides equal
SquareAll four sides equal AND diagonals equal

Derivation and Proof

Derivation of the Distance Formula:

The Distance Formula is derived directly from the Pythagoras Theorem applied to the Cartesian coordinate system.

Setup: Let P(x_1, y_1) and Q(x_2, y_2) be two points in the Cartesian plane. We want to find the distance PQ.

Step 1: Construction. Draw a horizontal line through P and a vertical line through Q (or vice versa). These lines intersect at a point R, forming a right-angled triangle PQR with the right angle at R.

The coordinates of R are (x_2, y_1) (it shares its x-coordinate with Q and its y-coordinate with P).

Step 2: Find the lengths of the two legs of the right triangle.

The horizontal leg PR has length = |x_2 - x_1| (the horizontal distance between P and R).

The vertical leg QR has length = |y_2 - y_1| (the vertical distance between Q and R).

Step 3: Apply the Pythagoras Theorem.

In the right triangle PQR with hypotenuse PQ:

PQ^2 = PR^2 + QR^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2

Note: We can drop the absolute value signs because squaring makes any negative value positive.

Step 4: Take the positive square root.

PQ = sqrt[(x_2 - x_1)^2 + (y_2 - y_1)^2]

Since distance is always positive, we take the positive square root.

Conclusion: The distance between P(x_1, y_1) and Q(x_2, y_2) is sqrt[(x_2 - x_1)^2 + (y_2 - y_1)^2].

Special Case (Distance from Origin): If P is the origin (0, 0) and Q is (x, y), then: PQ = sqrt[(x - 0)^2 + (y - 0)^2] = sqrt(x^2 + y^2).

Types and Properties

Problems involving the Distance Formula in Class 10 fall into several categories:

Type 1: Direct Calculation of Distance

Given two points, find the distance between them. This is the most basic application.

Type 2: Finding Distance from Origin

Find the distance of a given point from the origin O(0, 0).

Type 3: Checking Collinearity

Three points A, B, C are collinear (lie on the same straight line) if and only if AB + BC = AC (or any rearrangement where the two smaller distances add to the largest). Compute all three pairwise distances and check.

Type 4: Classifying Triangles

Given three vertices, compute the three side lengths using the distance formula and determine: (i) is it equilateral (all sides equal)? (ii) is it isosceles (two sides equal)? (iii) is it scalene (no sides equal)? (iv) is it right-angled (check the converse of Pythagoras)?

Type 5: Classifying Quadrilaterals

Given four vertices, compute all four sides and both diagonals to determine whether the quadrilateral is a parallelogram, rectangle, rhombus, square, or none of these.

Type 6: Finding Unknown Coordinates

Given the distance and one point, find the coordinates of another point. This leads to equations involving the distance formula.

Type 7: Equidistant Points

Find points equidistant from two or more given points. This often leads to the equation of a perpendicular bisector or a circle.

Methods

Method 1: Direct Distance Computation

Substitute the coordinates directly into d = sqrt[(x_2 - x_1)^2 + (y_2 - y_1)^2]. Simplify step by step.

Example: Distance between (3, -2) and (-1, 5): d = sqrt[(-1-3)^2 + (5-(-2))^2] = sqrt[16 + 49] = sqrt(65).

Method 2: Collinearity Check

For points A, B, C: compute AB, BC, and AC. If the sum of the two smaller distances equals the largest, the points are collinear.

Example: A(1,1), B(3,3), C(5,5). AB = sqrt(4+4) = 2sqrt(2). BC = sqrt(4+4) = 2sqrt(2). AC = sqrt(16+16) = 4sqrt(2). AB + BC = 4sqrt(2) = AC. Collinear.

Method 3: Triangle Classification

Compute all three side lengths. Check for equality (equilateral/isosceles). Check c^2 vs a^2 + b^2 for right angle.

Method 4: Quadrilateral Classification

Compute all four sides (AB, BC, CD, DA) and both diagonals (AC, BD). Compare to identify the type.

Method 5: Finding Unknown Coordinates

Set up the distance formula equation with the unknown coordinate and the given distance. Square both sides to eliminate the square root, then solve the resulting equation.

Example: Find x if the distance from (x, 3) to (2, 7) is 5. 5 = sqrt[(x-2)^2 + 16]. Squaring: 25 = (x-2)^2 + 16. (x-2)^2 = 9. x-2 = +/-3. x = 5 or x = -1.

Tips:

  • Always compute (x_2 - x_1) and (y_2 - y_1) separately before squaring to avoid arithmetic errors.
  • The order of points does not matter: d(P,Q) = d(Q,P).
  • When comparing distances, you can compare the squares (avoid square roots) to check equality or inequality.
  • Leave answers in surd form (e.g., sqrt(65)) unless asked to approximate.

Solved Examples

Example 1: Finding the Distance Between Two Points

Problem: Find the distance between the points A(2, 3) and B(6, 6).

Solution:

Step 1: d = sqrt[(6-2)^2 + (6-3)^2]

Step 2: d = sqrt[4^2 + 3^2] = sqrt[16 + 9] = sqrt(25) = 5

Answer: AB = 5 units.

Example 2: Distance from the Origin

Problem: Find the distance of the point P(-5, 12) from the origin.

Solution:

Step 1: d = sqrt[(-5)^2 + 12^2] = sqrt[25 + 144] = sqrt(169) = 13

Answer: The distance is 13 units.

Example 3: Checking Collinearity of Three Points

Problem: Show that the points A(1, -1), B(5, 2), and C(9, 5) are collinear.

Solution:

Step 1: AB = sqrt[(5-1)^2 + (2-(-1))^2] = sqrt[16 + 9] = sqrt(25) = 5.

Step 2: BC = sqrt[(9-5)^2 + (5-2)^2] = sqrt[16 + 9] = sqrt(25) = 5.

Step 3: AC = sqrt[(9-1)^2 + (5-(-1))^2] = sqrt[64 + 36] = sqrt(100) = 10.

Step 4: AB + BC = 5 + 5 = 10 = AC.

Since AB + BC = AC, the points A, B, C are collinear.

Answer: The points are collinear. B is the midpoint of AC.

Example 4: Proving a Triangle Is Isosceles

Problem: Show that the points A(0, 0), B(3, sqrt(3)), and C(-3, sqrt(3)) form an isosceles triangle.

Solution:

Step 1: AB = sqrt[(3-0)^2 + (sqrt(3)-0)^2] = sqrt[9 + 3] = sqrt(12) = 2sqrt(3).

Step 2: AC = sqrt[(-3-0)^2 + (sqrt(3)-0)^2] = sqrt[9 + 3] = sqrt(12) = 2sqrt(3).

Step 3: BC = sqrt[(-3-3)^2 + (sqrt(3)-sqrt(3))^2] = sqrt[36 + 0] = 6.

Step 4: Since AB = AC = 2sqrt(3) (two sides are equal), triangle ABC is isosceles with AB = AC.

Answer: Triangle ABC is isosceles.

Example 5: Proving a Quadrilateral Is a Square

Problem: Show that the points A(1, 2), B(4, 6), C(8, 3), and D(5, -1) form a square.

Solution:

Step 1: AB = sqrt[(4-1)^2 + (6-2)^2] = sqrt[9 + 16] = sqrt(25) = 5.

Step 2: BC = sqrt[(8-4)^2 + (3-6)^2] = sqrt[16 + 9] = sqrt(25) = 5.

Step 3: CD = sqrt[(5-8)^2 + (-1-3)^2] = sqrt[9 + 16] = sqrt(25) = 5.

Step 4: DA = sqrt[(1-5)^2 + (2-(-1))^2] = sqrt[16 + 9] = sqrt(25) = 5.

Step 5: All four sides are equal (= 5). The figure is at least a rhombus.

Step 6: Diagonal AC = sqrt[(8-1)^2 + (3-2)^2] = sqrt[49 + 1] = sqrt(50) = 5sqrt(2).

Step 7: Diagonal BD = sqrt[(5-4)^2 + (-1-6)^2] = sqrt[1 + 49] = sqrt(50) = 5sqrt(2).

Step 8: Diagonals are also equal (= 5sqrt(2)).

Since all four sides are equal AND both diagonals are equal, ABCD is a square.

Answer: ABCD is a square with side 5 and diagonal 5sqrt(2).

Example 6: Finding an Unknown Coordinate

Problem: Find the value of y if the distance between the points (2, y) and (5, 1) is sqrt(34).

Solution:

Step 1: sqrt(34) = sqrt[(5-2)^2 + (1-y)^2]

Step 2: Squaring: 34 = 9 + (1-y)^2

Step 3: (1-y)^2 = 25

Step 4: 1-y = 5 or 1-y = -5

Step 5: y = -4 or y = 6

Answer: y = -4 or y = 6.

Example 7: Equidistant Point on the X-Axis

Problem: Find a point on the x-axis that is equidistant from A(2, -5) and B(-2, 9).

Solution:

Step 1: A point on the x-axis has coordinates (x, 0) for some value x.

Step 2: Distance from (x, 0) to A(2, -5) = sqrt[(x-2)^2 + 25].

Step 3: Distance from (x, 0) to B(-2, 9) = sqrt[(x+2)^2 + 81].

Step 4: Setting them equal and squaring: (x-2)^2 + 25 = (x+2)^2 + 81.

Step 5: x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81.

Step 6: -4x + 29 = 4x + 85. So -8x = 56. x = -7.

Answer: The point is (-7, 0).

Example 8: Proving a Right-Angled Triangle

Problem: Prove that the points A(1, 7), B(4, 2), and C(-1, -1) form a right-angled isosceles triangle.

Solution:

Step 1: AB = sqrt[(4-1)^2 + (2-7)^2] = sqrt[9 + 25] = sqrt(34).

Step 2: BC = sqrt[(-1-4)^2 + (-1-2)^2] = sqrt[25 + 9] = sqrt(34).

Step 3: AC = sqrt[(-1-1)^2 + (-1-7)^2] = sqrt[4 + 64] = sqrt(68).

Step 4: AB = BC = sqrt(34). The triangle is isosceles.

Step 5: Check right angle: AB^2 + BC^2 = 34 + 34 = 68 = AC^2. By the converse of Pythagoras, angle B = 90 degrees.

Answer: Triangle ABC is a right-angled isosceles triangle with the right angle at B.

Example 9: Application: Shortest Distance Problem

Problem: A mobile tower is located at point A(3, 4). Two houses are located at B(7, 7) and C(11, 4). Which house is nearer to the tower, and by how much?

Solution:

Step 1: AB = sqrt[(7-3)^2 + (7-4)^2] = sqrt[16 + 9] = sqrt(25) = 5 units.

Step 2: AC = sqrt[(11-3)^2 + (4-4)^2] = sqrt[64 + 0] = 8 units.

Step 3: Since AB = 5 < AC = 8, house B is nearer to the tower.

Step 4: Difference = AC - AB = 8 - 5 = 3 units.

Answer: House B is nearer to the tower by 3 units.

Example 10: Distance and Perimeter of a Triangle

Problem: Find the perimeter of the triangle with vertices P(0, 0), Q(5, 0), and R(0, 12).

Solution:

Step 1: PQ = sqrt[(5-0)^2 + (0-0)^2] = 5.

Step 2: QR = sqrt[(0-5)^2 + (12-0)^2] = sqrt[25 + 144] = sqrt(169) = 13.

Step 3: PR = sqrt[(0-0)^2 + (12-0)^2] = 12.

Step 4: Perimeter = PQ + QR + PR = 5 + 13 + 12 = 30 units.

Note: This is a right triangle (5-12-13 triplet) with the right angle at P.

Answer: The perimeter is 30 units.

Real-World Applications

The Distance Formula has widespread applications across mathematics, science, and everyday life.

Navigation and GPS: Modern GPS systems calculate the distance between two locations using a version of the distance formula adapted for spherical coordinates. For short distances on a flat surface, the standard Cartesian distance formula provides accurate results. Navigation apps use this to calculate route distances, estimated travel times, and to find the nearest facilities.

Computer Graphics and Gaming: In video games and computer graphics, the distance formula is used millions of times per frame to calculate distances between objects, detect collisions, render lighting effects (distance from light source to surface), and manage game mechanics like proximity-based interactions.

Robotics: Robots use the distance formula to navigate their environment, measure distances to obstacles, and plan paths. In warehouse automation, robots calculate the shortest distance to their destination using coordinate-based distance computations.

Astronomy: Astronomers use extensions of the distance formula to calculate distances between stars, planets, and galaxies. In a flat (Euclidean) approximation of space, the 3D distance formula d = sqrt[(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2] gives the spatial distance between celestial objects.

Urban Planning: City planners use coordinate geometry and the distance formula to plan road networks, determine optimal locations for schools, hospitals, and fire stations (minimising the maximum distance to any residential area), and design public transportation routes.

Sports Analytics: In cricket, football, and basketball analytics, player positions are tracked using coordinates, and the distance formula calculates pass distances, sprint lengths, and coverage areas. Heat maps of player movement are generated using accumulated distance computations.

Machine Learning: Many machine learning algorithms (such as K-nearest neighbours, K-means clustering, and support vector machines) use the Euclidean distance formula to measure similarity between data points. The distance between feature vectors determines classification and clustering outcomes.

Key Points to Remember

  • The Distance Formula: d = sqrt[(x_2 - x_1)^2 + (y_2 - y_1)^2] gives the distance between two points in the Cartesian plane.
  • It is derived directly from the Pythagoras Theorem by constructing a right triangle with horizontal and vertical legs.
  • Distance from origin: d = sqrt(x^2 + y^2).
  • Distance is always non-negative and symmetric: d(P,Q) = d(Q,P).
  • Collinearity check: points A, B, C are collinear if AB + BC = AC.
  • For triangle classification, compute all three sides and check for equilateral, isosceles, or right-angled conditions.
  • For quadrilateral classification, compute all four sides and both diagonals.
  • When finding unknown coordinates, set up the distance equation and solve algebraically.
  • For equidistant problems, set two distances equal and solve for the unknown.
  • In CBSE board exams, distance formula problems carry 2-4 marks. Common question types: finding distances, proving triangle/quadrilateral types, collinearity, and finding unknown coordinates.

Practice Problems

  1. Find the distance between the points (-3, 4) and (5, -2).
  2. Show that the points (1, 5), (2, 3), and (4, -1) are collinear.
  3. Prove that the points (0, 0), (0, 5), (5, 5), and (5, 0) form a square.
  4. Find the point on the y-axis equidistant from (5, -2) and (-3, 2).
  5. The distance between (a, 2) and (3, 4) is sqrt(8). Find the value of a.
  6. Show that the triangle with vertices (4, 3), (-1, -1), and (8, -1) is an isosceles right triangle.
  7. Find the perimeter of the quadrilateral with vertices (0, 0), (5, 0), (7, 3), and (2, 3).
  8. A circle has centre (2, 3) and passes through (5, 7). Find its radius.

Frequently Asked Questions

Q1. What is the Distance Formula?

The Distance Formula gives the distance between two points P(x_1, y_1) and Q(x_2, y_2) as d = sqrt[(x_2 - x_1)^2 + (y_2 - y_1)^2]. It is derived from the Pythagoras Theorem applied to the coordinate plane.

Q2. Does the order of points matter in the Distance Formula?

No. Since the differences are squared, (x_2 - x_1)^2 = (x_1 - x_2)^2, so d(P,Q) = d(Q,P). The distance is the same regardless of which point you call first.

Q3. How do you check if three points are collinear?

Compute the distances AB, BC, and AC between all three pairs. If the sum of the two smaller distances equals the largest, the points are collinear. For example, if AB + BC = AC, then B lies on segment AC.

Q4. How do you prove a quadrilateral is a square using the Distance Formula?

Compute all four sides (AB, BC, CD, DA) and both diagonals (AC, BD). If all four sides are equal AND both diagonals are equal, the quadrilateral is a square. If only all sides are equal (diagonals not necessarily equal), it is a rhombus.

Q5. What is the distance of a point from the origin?

The distance of point (x, y) from the origin (0, 0) is sqrt(x^2 + y^2). This follows directly from the distance formula with one point as the origin.

Q6. Can the Distance Formula give a negative answer?

No. Since we take the positive square root of a sum of squares (which are always non-negative), the distance is always non-negative. It equals zero only when the two points are the same.

Q7. How is the Distance Formula related to the Pythagoras Theorem?

The Distance Formula IS the Pythagoras Theorem applied to coordinates. The horizontal distance |x_2 - x_1| and vertical distance |y_2 - y_1| are the two legs of a right triangle, and the actual distance between the points is the hypotenuse.

Q8. Can the Distance Formula be extended to three dimensions?

Yes. In 3D, the distance between (x_1, y_1, z_1) and (x_2, y_2, z_2) is d = sqrt[(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2]. This is studied in higher classes but follows the same principle.

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