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ICSE • Class 9 • Mathematics • Ch 24
Estimated Time: 45 Mins
Study Progress: In Progress

Distance Formula

In ICSE Class 9 Mathematics, "Distance Formula" represents the metric culmination of coordinate geometry, providing the universal algebraic formula for calculating the exact Euclidean distance between any two points in the Cartesian plane. Given two points $P(x_1, y_1)$ and $Q(x_2, y_2)$, the distance $d(P, Q)$ is derived directly by constructing a right-angled triangle $PQR$ (with $R(x_2, y_1)$) and applying the Pythagoras Theorem ($PQ^2 = PR^2 + QR^2$): $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. As an immediate corollary, the distance of any point $P(x, y)$ from the origin $O(0, 0)$ is given by $OP = \sqrt{x^2 + y^2}$. The chapter systematically trains students in applying the distance formula to geometric classifications: (1) Testing collinearity of three points $A, B, C$ (collinear if and only if the sum of the two smaller distances equals the largest distance: $AB + BC = AC$); (2) Classifying triangles as equilateral (all 3 distances equal), isosceles (2 distances equal), scalene (all unequal), or right-angled (satisfying $a^2 + b^2 = c^2$); (3) Classifying quadrilaterals: Parallelogram (opposite sides equal), Rhombus (all 4 sides equal, diagonals unequal), Rectangle (opposite sides equal and diagonals equal), and Square (all 4 sides equal and diagonals equal); (4) Finding equidistant points (such as the circumcenter of a triangle); and (5) Solving algebraic equations to find unknown coordinates.

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How Smartphone Uber and Google Maps Calculate Your Travel Distance in Microseconds

When you open Uber or Google Maps on your phone, you see your current location as a glowing blue dot at coordinates $(x_1, y_1)$ and your destination as a pin at $(x_2, y_2)$. In less than a thousandth of a second, the app calculates the exact straight-line distance, estimates your fare, and finds the nearest driver. How does a computer chip instantly compute physical distance from satellite latitude and longitude coordinates? It uses the Distance Formula! By creating an invisible right-angled triangle across the globe—with the horizontal difference in longitude as the base $(x_2 - x_1)$ and the vertical difference in latitude as the height $(y_2 - y_1)$—the Pythagorean theorem computes the hypotenuse: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$! How do forensic scientists use this formula to reconstruct crime scenes, and how do civil engineers prove whether four surveyed stakes form a true square? Let us master the distance formula!

Why This Chapter Matters

The distance formula is foundational for GPS satellite navigation, machine learning clustering algorithms ($k$-Nearest Neighbors), collision detection in video games, robotics, and architectural surveying.

Before You Begin (Prerequisites)

  • Cartesian coordinate plotting from Chapter 22.
  • The Pythagoras Theorem ($a^2 + b^2 = c^2$) from Chapter 12.

What You Will Learn (Core Objectives)

  • State, derive, and apply the Euclidean Distance Formula between two points $(x_1, y_1)$ and $(x_2, y_2)$.
  • Calculate the distance of any point $(x, y)$ from the origin $O(0, 0)$.
  • Prove collinearity of three given points using distance addition ($AB + BC = AC$).
  • Classify triangles (equilateral, isosceles, right-angled) from given vertex coordinates.
  • Prove whether four given coordinates form a parallelogram, rectangle, rhombus, or square.

Chapter Roadmap & Progression

1 1. Derivation of the Distance Formu...
2 2. Geometric Applications of the Di...
3 3. Worked ICSE Problem Archetypes

Complete Concept Guide (100% Curriculum Coverage)

1. Derivation of the Distance Formula

Formal Derivation
Theorem:

The distance $d$ between any two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ in the Cartesian plane is:

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

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be two given points. Draw $PL$ and $QM$ perpendicular to the $X$-axis. From $P$, draw $PR \perp QM$ meeting $QM$ at $R$.

  • The coordinates of $R$ are $(x_2, y_1)$.
  • Horizontal leg $PR = LM = OM - OL = x_2 - x_1$.
  • Vertical leg $QR = QM - RM = QM - PL = y_2 - y_1$.

In right-angled triangle $\triangle PRQ$, by the Pythagoras Theorem:

$$PQ^2 = PR^2 + QR^2$$ $$PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$$ $$\mathbf{PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}} \quad \blacksquare$$
Distance from the Origin:

If $P$ is $(x, y)$ and $O$ is $(0, 0)$:

$$\mathbf{OP = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}}$$

2. Geometric Applications of the Distance Formula

Geometric Classifications
A. Collinearity Test:

Three points $A, B, C$ are collinear if and only if the sum of the lengths of two segments equals the third:

$$\mathbf{AB + BC = AC}$$
B. Classification of Triangles:
  • Equilateral Triangle: $AB = BC = CA$.
  • Isosceles Triangle: Any two sides equal (e.g., $AB = AC \neq BC$).
  • Right-Angled Triangle: Square of longest side equals sum of squares of other two: $AB^2 + BC^2 = AC^2$.
C. Classification of Quadrilaterals:
Quadrilateral TypeSides ConditionDiagonals Condition
ParallelogramOpposite sides equal ($AB = CD$ and $BC = DA$)Diagonals unequal ($AC \neq BD$)
RectangleOpposite sides equal ($AB = CD$ and $BC = DA$)Diagonals EQUAL ($AC = BD$)
RhombusAll 4 sides equal ($AB = BC = CD = DA$)Diagonals unequal ($AC \neq BD$)
SquareAll 4 sides equal ($AB = BC = CD = DA$)Diagonals EQUAL ($AC = BD$)

3. Worked ICSE Problem Archetypes

Exemplary Solutions
Problem 1: Find the distance between the points $A(-3, 6)$ and $B(2, -6)$.

Solution:

Here, $x_1 = -3, y_1 = 6$ and $x_2 = 2, y_2 = -6$.

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{[2 - (-3)]^2 + [-6 - 6]^2}$$ $$= \sqrt{(5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = \mathbf{13\text{ units}}$$
Problem 2: Prove that the points $A(1, 7), B(4, 2), C(-1, -1), D(-4, 4)$ are the vertices of a square.

Solution:

1. Calculate the lengths of all 4 sides:

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

Since $AB = BC = CD = DA = \sqrt{34}$, all four sides are equal (it is at least a rhombus).

2. Calculate the lengths of both diagonals:

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

Since the diagonals are equal ($AC = BD = \sqrt{68}$), $ABCD$ is strictly a SQUARE. $\blacksquare$

Key Formulas, Identities & Theorems

Euclidean Distance Formula
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Distance between any two points (x1, y1) and (x2, y2).
Distance from Origin
$$d = \sqrt{x^2 + y^2}$$
Distance of point (x, y) from (0, 0).
Collinearity Condition
AB + BC = AC
Sum of two segments equals third segment.

Mathematics: Derivation of Distance Formula via Pythagorean Right Triangle

Distance Formula: Pythagorean Triangle Construction in Cartesian Plane X Y O P (x₁, y₁) Q (x₂, y₂) R (x₂, y₁) Base PR = (x₂ - x₁) Height QR = (y₂ - y₁) Distance d x₁ x₂ Pythagoras in ΔPRQ: d² = (x₂ - x₁)² + (y₂ - y₁)²

Chapter Summary & 10 Key Takeaways

Takeaway 1
The distance between two points P(x1, y1) and Q(x2, y2) is d = √[(x2 - x1)^2 + (y2 - y1)^2].
Takeaway 2
The distance of point P(x, y) from the origin (0, 0) is d = √(x^2 + y^2).
Takeaway 3
Distance is strictly a scalar non-negative quantity: d ≥ 0.
Takeaway 4
Three points A, B, C are collinear if and only if AB + BC = AC.
Takeaway 5
A triangle is equilateral if all 3 side lengths are equal.
Takeaway 6
A triangle is right-angled if the square of the longest side equals the sum of squares of the other two.
Takeaway 7
A quadrilateral is a rectangle if opposite sides are equal and diagonals are equal.
Takeaway 8
A quadrilateral is a rhombus if all four sides are equal and diagonals are unequal.
Takeaway 9
A quadrilateral is a square if all four sides are equal and diagonals are equal.
Takeaway 10
Equidistant points from two given points lie on their perpendicular bisector.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Find the distance of the point $P(-5, 12)$ from the origin $O(0, 0)$.
Reveal Answer & Explanation
Answer: • Using the distance from the origin formula $d = \sqrt{x^2 + y^2}$:
$$d = \sqrt{(-5)^2 + (12)^2} = \sqrt{25 + 144} = \sqrt{169} = \mathbf{13\text{ units}}$$
d = √((-5)^2 + 12^2) = √(25 + 144) = 13.
2
Check whether the points $A(1, 5), B(2, 3)$, and $C(-2, -11)$ are collinear.
Reveal Answer & Explanation
Answer:

• Calculate the distances between all three pairs of points:
1. $AB = \sqrt{(2 - 1)^2 + (3 - 5)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$
2. $BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2} = \sqrt{(-4)^2 + (-14)^2} = \sqrt{16 + 196} = \sqrt{212} = 2\sqrt{53}$
3. $AC = \sqrt{(-2 - 1)^2 + (-11 - 5)^2} = \sqrt{(-3)^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265}$
• Check if the sum of any two distances equals the third:
$AB + BC = \sqrt{5} + 2\sqrt{53} \approx 2.236 + 14.56 = 16.796$
$AC = \sqrt{265} \approx 16.279$
• Since $AB + BC \neq AC$ (and no other combination sums to the third), the points are NOT collinear.


Find AB = √5, BC = √212, AC = √265. Check if AB + BC = AC.
3
Find the value of $y$ if the distance between the points $A(2, -3)$ and $B(10, y)$ is $10\text{ units}$.
Reveal Answer & Explanation
Answer:

• Using the distance formula:

$$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$$


$$10^2 = (10 - 2)^2 + [y - (-3)]^2$$


$$100 = 8^2 + (y + 3)^2 = 64 + (y + 3)^2$$


$$(y + 3)^2 = 100 - 64 = 36$$


• Taking square roots on both sides:

$$y + 3 = \pm 6$$


• Solving both cases:
- Case 1: $y + 3 = 6 \implies \mathbf{y = 3}$
- Case 2: $y + 3 = -6 \implies \mathbf{y = -9}$
• Values of $y$: $\mathbf{y = 3 \text{ or } y = -9}$.


(y + 3)^2 = 100 - 64 = 36. y + 3 = ±6 -> y = 3 or y = -9.
4
Show that the points $A(7, 10), B(-2, 5)$, and $C(3, -4)$ are the vertices of an isosceles right-angled triangle.
Reveal Answer & Explanation
Answer:

• Calculate the squares of all three sides:
1. $AB^2 = (-2 - 7)^2 + (5 - 10)^2 = (-9)^2 + (-5)^2 = 81 + 25 = 106$
2. $BC^2 = [3 - (-2)]^2 + (-4 - 5)^2 = 5^2 + (-9)^2 = 25 + 81 = 106$
3. $CA^2 = (7 - 3)^2 + [10 - (-4)]^2 = 4^2 + 14^2 = 16 + 196 = 212$
• Isosceles Check:
$AB^2 = BC^2 = 106 \implies AB = BC = \sqrt{106}$. Two sides are equal, so the triangle is isosceles.
• Right-Angled Check (Converse Pythagoras):

$$AB^2 + BC^2 = 106 + 106 = 212 = CA^2$$


• By the Converse of the Pythagoras Theorem, $\angle B = 90^\circ$.
• Hence, $\triangle ABC$ is an isosceles right-angled triangle. $\blacksquare$


AB^2 = 106, BC^2 = 106, CA^2 = 212. AB = BC and AB^2 + BC^2 = CA^2.
5
Find a point on the $X$-axis which is equidistant from $A(2, -5)$ and $B(-2, 9)$.
Reveal Answer & Explanation
Answer: • Any point on the $X$-axis has coordinates $P(x, 0)$.
• Since $P$ is equidistant from $A$ and $B$, $PA = PB \implies PA^2 = PB^2$.
$$PA^2 = (x - 2)^2 + [0 - (-5)]^2 = (x - 2)^2 + 25$$
$$PB^2 = [x - (-2)]^2 + (0 - 9)^2 = (x + 2)^2 + 81$$
• Equating $PA^2 = PB^2$:
$$(x - 2)^2 + 25 = (x + 2)^2 + 81$$
$$(x^2 - 4x + 4) + 25 = (x^2 + 4x + 4) + 81$$
$$-4x + 29 = 4x + 85 \implies 8x = 29 - 85 = -56 \implies x = -7$$
• The required point is $\mathbf{(-7, 0)}$.
Set P(x, 0). Solve (x - 2)^2 + 25 = (x + 2)^2 + 81 to find x = -7.
6
What type of quadrilateral has vertices $A(0, -1), B(2, 1), C(0, 3), D(-2, 1)$? Justify.
Reveal Answer & Explanation
Answer:

• Calculate sides:
- $AB = \sqrt{(2-0)^2 + (1 - (-1))^2} = \sqrt{4 + 4} = \sqrt{8}$
- $BC = \sqrt{(0-2)^2 + (3-1)^2} = \sqrt{4 + 4} = \sqrt{8}$
- $CD = \sqrt{(-2-0)^2 + (1-3)^2} = \sqrt{4 + 4} = \sqrt{8}$
- $DA = \sqrt{(0 - (-2))^2 + (-1 - 1)^2} = \sqrt{4 + 4} = \sqrt{8}$
All four sides are equal ($AB = BC = CD = DA = \sqrt{8}$).
• Calculate diagonals:
- $AC = \sqrt{(0-0)^2 + (3 - (-1))^2} = \sqrt{16} = 4$
- $BD = \sqrt{(-2-2)^2 + (1-1)^2} = \sqrt{16} = 4$
• Both diagonals are equal ($AC = BD = 4$).
• A quadrilateral with four equal sides and equal diagonals is a SQUARE.


All 4 sides = √8 and both diagonals = 4. It is a square.
7
Find the perimeter of $\triangle ABC$ whose vertices are $A(0, 4), B(0, 0), C(3, 0)$.
Reveal Answer & Explanation
Answer:

• $B(0, 0)$ is the origin.
• $AB$ lies along the $Y$-axis: length $= |4 - 0| = 4\text{ units}$.
• $BC$ lies along the $X$-axis: length $= |3 - 0| = 3\text{ units}$.
• Hypotenuse $AC = \sqrt{(3 - 0)^2 + (0 - 4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ units}$.
• Perimeter:

$$P = AB + BC + AC = 4 + 3 + 5 = \mathbf{12\text{ units}}$$


Sides are 4, 3, and hypotenuse 5. Perimeter = 4 + 3 + 5 = 12 units.
8
Explain how the distance formula is derived from the Pythagoras Theorem.
Reveal Answer & Explanation
Answer: • To find distance between $P(x_1, y_1)$ and $Q(x_2, y_2)$, a third point $R(x_2, y_1)$ is constructed such that $PR$ is horizontal and $QR$ is vertical.
• Since the coordinate axes are perpendicular, $\triangle PRQ$ is a right-angled triangle with $\angle R = 90^\circ$.
• The horizontal leg is $PR = |x_2 - x_1|$ and the vertical leg is $QR = |y_2 - y_1|$.
• Applying the Pythagoras Theorem to $\triangle PRQ$:
$$PQ^2 = PR^2 + QR^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$$
Taking the positive square root gives the Euclidean distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Construct horizontal segment |x2 - x1| and vertical segment |y2 - y1| forming right triangle; hypotenuse is d.
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