The Distance Formula: One Equation for Every Distance — Class 9 Mathematics Chapter 1

🖼 Image PendingA wide cinematic vista: a single elegant equation written in golden light spans across a starry sky, with light beams flowing into it from many pairs of points scattered across a coordinate plane below — symbolising one formula that captures every possible distance

AI Generation Prompt

Ultra-wide cinematic banner (16:5 ratio). A vast starry navy sky with a single elegant mathematical equation written in glowing golden light stretching horizontally across the upper third: d = √((x₂−x₁)² + (y₂−y₁)²). Below, a faint Cartesian coordinate plane stretches into the horizon, with many faint pairs of points scattered across it — each pair connected by a soft beam of light that flows up and into the equation, symbolising that this single formula contains every possible distance. Painterly cinematic illustration of mathematical wonder. Dark background. No text in foreground other than the equation itself.

✦Think About It

On the previous page you found three distances $AD$ , $DM$ , $MA$ — each by drawing a fresh right triangle and computing its hypotenuse. It worked, but it was a lot of separate steps for what feels like one operation: find the distance between two points.

Is there a single formula — one equation, applicable to any two points $(x_1, y_1)$ and $(x_2, y_2)$ — that gives the distance immediately?

Look back at the three calculations on Page 9. Each one had the same structure: $\sqrt{(\text{horizontal leg})^2 + (\text{vertical leg})^2}$ . The legs were just $|x_2 - x_1|$ and $|y_2 - y_1|$ . Can you write this once for any pair?

✦भास्कराचार्य — लीलावती

From the Līlāvatī of Bhāskara II (1150 CE)

कोटिवर्गयुतेन भुजवर्गेण विभक्तेन

कर्णवर्गः समायाति इति शुल्बस्य निश्चयः॥

(koṭi-varga-yutena bhuja-vargeṇa vibhaktena / karṇa-vargaḥ samāyāti iti śulbasya niścayaḥ)

'भुजा का वर्ग और कोटि का वर्ग जोड़ने पर कर्ण का वर्ग बनता है — यही शुल्ब-शास्त्र का निश्चय है।'

'The square of the base added to the square of the perpendicular gives the square of the hypotenuse — this is the conclusion of the Śulba.'

Bhāskarāchārya, writing in 1150 CE, restated Baudhāyana's 800 BCE result in a single Sanskrit verse — and added that this is "niścaya", the settled conclusion. The distance formula on this page is exactly this verse, written in modern algebraic notation. The mathematics of distance has not changed for 2,800 years; only the symbols have.

From three triangles to one formula

On the previous page, finding $AD$ , $DM$ , $MA$ each followed exactly the same recipe:

Find the horizontal separation: $|x_2 - x_1|$ .
Find the vertical separation: $|y_2 - y_1|$ .
Apply Baudhāyana–Pythagoras: distance² = (horizontal)² + (vertical)².
Take the square root.

Let me write this with general coordinates. Suppose the two points are $(x_1, y_1)$ and $(x_2, y_2)$ . Then:

Horizontal separation: $|x_2 - x_1|$
Vertical separation: $|y_2 - y_1|$
Distance² = $(x_2 - x_1)^2 + (y_2 - y_1)^2$
Distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

This last line is the distance formula. Notice an important simplification: the absolute-value bars disappear when we square the differences, because $(x_2 - x_1)^2 = |x_2 - x_1|^2$ for any real numbers. Squaring kills the sign. So we don't need to worry about which point we call $(x_1, y_1)$ and which one we call $(x_2, y_2)$ — the answer is the same.

🖼 Image PendingA green graph-paper Cartesian plane. Two points are marked: A(x₁, y₁) at upper-left and D(x₂, y₂) at lower-right. A horizontal dashed line runs from A's x-coordinate at y₂ height to D — its length labelled (x₂ - x₁). A vertical dashed line runs from A down to its foot at (x₁, y₂) — labelled (y₂ - y₁) but using y-axis convention. The diagonal segment from A to D is labelled with the distance formula expression √((x₂-x₁)² + (y₂-y₁)²). The right-angle marker is shown at the corner.

AI Generation Prompt

Cartesian plane diagram on green graph paper over a dark background. Two points labelled A(x₁, y₁) at the upper-left and D(x₂, y₂) at the lower-right are connected by a thick green diagonal segment. A right triangle is constructed with two dashed thin legs: one horizontal from A's x-coordinate level to D, labelled '(x₂ − x₁)' below; one vertical from A down to the foot at (x₁, y₂), labelled '(y₂ − y₁)' to the side. A small right-angle marker (square) is shown at the corner where the legs meet. The diagonal hypotenuse is labelled in elegant text: '√((x₂ − x₁)² + (y₂ − y₁)²)'. Style: clean educational mathematics illustration, generic-coordinate notation. Dark background, orange accent labels, clean technical illustration style.

📸 Fig. 1.7-1.8 — The general distance formula visualised. Two points $A(x_1, y_1)$ and $D(x_2, y_2)$ form the hypotenuse; the legs are $|x_2 - x_1|$ and $|y_2 - y_1|$.

Re-deriving the Page-9 results from the formula

Loading simulator…

Logical ReasoningLevel 3 · Analysis

In the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , suppose I swap $(x_1, y_1)$ and $(x_2, y_2)$ — i.e., I rename point 1 as point 2 and vice versa. Does the answer change?

The Distance Formula — Memorise This One Equation

For any two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ in the plane:

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

Special case — distance from the origin:

$d(O, P) = \sqrt{x_1^2 + y_1^2}$

Properties to remember:

$d(P, Q) \ge 0$ always — distance is non-negative.
$d(P, Q) = 0$ if and only if $P = Q$ .
$d(P, Q) = d(Q, P)$ — distance is symmetric.
The formula works in all four quadrants without modification — squaring handles all signs.

Practice Set — Twelve Problems for Drill

Cover the answers and solve each. Aim for accuracy first, speed second.

Find the distance between:

$(0, 0)$ and $(3, 4)$
$(2, 3)$ and $(5, 7)$
$(-1, -1)$ and $(2, 3)$
$(7, 0)$ and $(0, 24)$
$(-5, -12)$ and $(0, 0)$
$(6, -8)$ and $(0, 0)$
$(-2, 3)$ and $(4, 11)$
$(0, 0)$ and $(\sqrt{3}, 1)$
$(1, 1)$ and $(4, 5)$
$(a, 0)$ and $(0, b)$
$(2, 3)$ and $(2, -7)$ (same x)
$(p, q)$ and $(p + 5, q + 12)$

Answers: 1. 5 · 2. 5 · 3. 5 · 4. 25 · 5. 13 · 6. 10 · 7. 10 · 8. 2 · 9. 5 · 10. $\sqrt{a^2 + b^2}$ · 11. 10 (vertical segment) · 12. 13 (a 5-12-13 triangle in disguise).

Self-check rule: If your answer is negative, you've forgotten that distance is non-negative — recheck your squaring. If your answer is larger than the simple sum $|x_2 - x_1| + |y_2 - y_1|$ , it's wrong — the diagonal of a right triangle is always shorter than the sum of its legs.

Bridging Science and Society — Where the Distance Formula Lives Today

The distance formula is one of the most heavily-used pieces of mathematics in the modern world. Every minute, billions of computations of this exact formula happen on devices around the planet:

Ready to Go Beyond

Closely related formulas you'll meet later:

Quick Check

Q1.Find the distance between $(2, 3)$ and $(5, 7)$ .

AI Generation Prompt

✦Think About It

Is there a single formula — one equation, applicable to any two points $(x_1, y_1)$ and $(x_2, y_2)$ — that gives the distance immediately?

✦भास्कराचार्य — लीलावती

From the Līlāvatī of Bhāskara II (1150 CE)

कोटिवर्गयुतेन भुजवर्गेण विभक्तेन

कर्णवर्गः समायाति इति शुल्बस्य निश्चयः॥

(koṭi-varga-yutena bhuja-vargeṇa vibhaktena / karṇa-vargaḥ samāyāti iti śulbasya niścayaḥ)

'The square of the base added to the square of the perpendicular gives the square of the hypotenuse — this is the conclusion of the Śulba.'

From three triangles to one formula

On the previous page, finding $AD$ , $DM$ , $MA$ each followed exactly the same recipe:

Find the horizontal separation: $|x_2 - x_1|$ .
Find the vertical separation: $|y_2 - y_1|$ .
Apply Baudhāyana–Pythagoras: distance² = (horizontal)² + (vertical)².
Take the square root.

Let me write this with general coordinates. Suppose the two points are $(x_1, y_1)$ and $(x_2, y_2)$ . Then:

Horizontal separation: $|x_2 - x_1|$
Vertical separation: $|y_2 - y_1|$
Distance² = $(x_2 - x_1)^2 + (y_2 - y_1)^2$
Distance = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

AI Generation Prompt

📸 Fig. 1.7-1.8 — The general distance formula visualised. Two points $A(x_1, y_1)$ and $D(x_2, y_2)$ form the hypotenuse; the legs are $|x_2 - x_1|$ and $|y_2 - y_1|$.

Re-deriving the Page-9 results from the formula

Loading simulator…

Logical ReasoningLevel 3 · Analysis

In the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , suppose I swap $(x_1, y_1)$ and $(x_2, y_2)$ — i.e., I rename point 1 as point 2 and vice versa. Does the answer change?

Bridging Science and Society — Where the Distance Formula Lives Today

The distance formula is one of the most heavily-used pieces of mathematics in the modern world. Every minute, billions of computations of this exact formula happen on devices around the planet:

Ready to Go Beyond

Closely related formulas you'll meet later:

Quick Check

Q1.Find the distance between $(2, 3)$ and $(5, 7)$ .