Distance Between Two Points: From Axes to Diagonals — Class 9 Mathematics Chapter 1

🖼 Image PendingA wide cinematic visualisation: a single straight line of golden light cuts diagonally across a deep starry sky, with two faint pillars of light extending from its endpoints — one horizontal, one vertical — meeting at a right angle below it, evoking the secret triangle hidden inside any diagonal

AI Generation Prompt

Ultra-wide cinematic banner (16:5 ratio). A single straight line of bright golden light cuts diagonally across a deep, starry navy sky. From each endpoint of this diagonal, a faint pillar of light extends — one horizontal, one vertical — meeting at a right angle just below the diagonal, forming a perfect right triangle visible only as overlapping beams of light. The image conveys: the distance between two points in a plane is just the hypotenuse of a hidden right triangle. Painterly cinematic illustration in the style of mathematical wonder. Dark background. No text, no labels.

✦Think About It

Two points sit on the same horizontal line at $(2, 5)$ and $(7, 5)$ . The distance between them is 5 units — you found this on a previous page just by subtracting the x-coordinates.

Now let me move one of those points slightly: $(2, 5)$ stays, but the other becomes $(7, 1)$ . The two points are no longer on the same line. What is the distance between them now?

Is it still 5? More? Less? And how would you compute it without a ruler?

If you draw a horizontal line from one point and a vertical line from the other, you build a right triangle. The diagonal you want is its hypotenuse.

✦बौधायन शुल्बसूत्र — १.१२

The Baudhāyana–Pythagoras Theorem

दीर्घचतुरस्रस्य अक्ष्णयारज्जुः

पार्श्वमानी तिर्यङ्मानी च

यत् पृथग् भूते कुरुतः तदुभयं करोति॥

(dīrghacaturasrasya akṣṇayārajjuḥ / pārśvamānī tiryaṅmānī ca / yat pṛthag bhūte kurutaḥ tadubhayaṃ karoti)

'किसी आयत के विकर्ण-पर बना वर्ग, उसकी दोनों भुजाओं पर बने वर्गों के योग के बराबर होता है।'

'The square on the diagonal of a rectangle equals the sum of the squares on the two sides.'

This is the Pythagoras theorem. Baudhāyana wrote it down in the Śulba-sūtra around 800 BCE — over 300 years before Pythagoras. In modern symbols: $a^2 + b^2 = c^2$ . You will use this verse, exactly, on every problem on this page.

Easy distances first: along an axis or parallel to it

On previous pages you found distances between points whenever the segment between them was parallel to one of the axes. The rule was simple subtraction:

Two points $(x_1, y)$ and $(x_2, y)$ on the same horizontal line (same y-coordinate): the distance is $|x_2 - x_1|$ .
Two points $(x, y_1)$ and $(x, y_2)$ on the same vertical line (same x-coordinate): the distance is $|y_2 - y_1|$ .

For example, in Reiaan's room (Page 5), the room door $D_1(7.5, 0)$ to $R_1(11.5, 0)$ is horizontal — its width is $|11.5 - 7.5| = 4$ ft. Easy.

But the segment from $(2, 5)$ to $(7, 1)$ is not parallel to either axis. Subtracting the x-coordinates gives 5, subtracting the y-coordinates gives 4 — neither of these is the distance between the two points. They are something else: the legs of a right triangle whose hypotenuse is the distance we want.

This is the moment Baudhāyana's theorem becomes a tool of coordinate geometry.

Triangle ADM: the worked example

Consider three points in Quadrant I:

$A = (3, 4)$
$D = (7, 1)$
$M = (9, 6)$

Draw segments $AD$ , $DM$ , $MA$ on graph paper. The result is a triangle entirely inside Quadrant I. Triangle ADM is acute-angled (all three angles are less than 90°). Now we ask the obvious question: how long are its sides $AD$ , $DM$ , $MA$ ?

None of these segments is parallel to an axis. Plain subtraction will not give us any of the three side lengths directly. But we can build a right triangle for each side, where the side is the hypotenuse and the legs are parallel to the axes.

🖼 Image PendingA green graph-paper Cartesian plane on a dark background. Three points are plotted in Quadrant I: A = (3, 4), D = (7, 1), M = (9, 6). Each is labelled with its coordinates. The three points are connected by line segments AD, DM, MA, forming a triangle that lies entirely in the upper-right region. The x-axis is labelled with integers from −1 to 10; the y-axis from 0 to 6.

AI Generation Prompt

Cartesian plane diagram on green graph paper over a dark background. Two thick green axes meet at the origin. Tick marks at every integer from −1 to +10 on x-axis and from 0 to +6 on y-axis. Three points plotted as small dark dots, each labelled with both name and coordinates: A(3, 4), D(7, 1), M(9, 6). Three green line segments connect them to form an acute-angled triangle ADM in Quadrant I. Style: clean educational mathematics illustration. Dark background, orange accent labels, clean technical illustration style.

📸 Fig. 1.6 — Triangle ADM with $A(3, 4)$, $D(7, 1)$, $M(9, 6)$, drawn entirely inside Quadrant I.

Focus on side $AD$ first. The endpoints are $A(3, 4)$ and $D(7, 1)$ . To find $AD$ :

Drop a vertical line from $A$ down to the level of $D$ (i.e. to $y = 1$ ). Call the foot $C$ . Then $C = (3, 1)$ .
Now we have three points: $A(3, 4)$ , $C(3, 1)$ , $D(7, 1)$ . Notice that $\angle ACD = 90°$ — $C$ is directly below $A$ and directly left of $D$ , so the segment $AC$ is vertical and the segment $CD$ is horizontal. Triangle $ACD$ is right-angled at $C$ .
Now compute the two legs by simple subtraction:
- $CD$ = horizontal distance = $|7 - 3| = 4$
- $AC$ = vertical distance = $|4 - 1| = 3$
By Baudhāyana–Pythagoras: $AD^2 = AC^2 + CD^2 = 3^2 + 4^2 = 9 + 16 = 25$ , so $AD = \sqrt{25} = 5$ .

The diagonal distance from $A(3, 4)$ to $D(7, 1)$ is exactly 5 units. This is the same 3-4-5 right triangle Baudhāyana would have used to lay out a perfect right angle on the ground 2,800 years ago — and it works because of the same theorem he proved.

The pattern is so important it deserves a name. The distance between two points in a plane is the hypotenuse of the right triangle whose legs are parallel to the axes.

Loading simulator…

Worked Example 1Find AD

SOLVED

Find the length of the segment from $A(3, 4)$ to $D(7, 1)$ .

Solution

Step 1 — Find the legs of the right triangle.

Horizontal leg = $|x_D - x_A| = |7 - 3| = 4$
Vertical leg = $|y_A - y_D| = |4 - 1| = 3$

Step 2 — Apply Baudhāyana–Pythagoras.

$AD^2 = 4^2 + 3^2 = 16 + 9 = 25$

Step 3 — Take the square root.

$AD = \sqrt{25} = 5$

Answer: $AD = 5$ units.

Logical ReasoningLevel 3 · Analysis

Notice that for $A(3, 4)$ and $D(7, 1)$ , both simple coordinate subtractions give: $|7 - 3| = 4$ and $|4 - 1| = 3$ . Neither of these is the distance $AD = 5$ .

Why is the distance not just one of these subtractions, the way it was for points on a horizontal or vertical line?

Worked Example 2Find DM

SOLVED

Find the length of the segment from $D(7, 1)$ to $M(9, 6)$ .

Solution

Step 1 — Find the legs.

Horizontal leg = $|x_M - x_D| = |9 - 7| = 2$
Vertical leg = $|y_M - y_D| = |6 - 1| = 5$

Step 2 — Apply Baudhāyana–Pythagoras.

$DM^2 = 2^2 + 5^2 = 4 + 25 = 29$

Step 3 — Take the square root.

$DM = \sqrt{29} \approx 5.39$

Answer: $DM = \sqrt{29}$ units (≈ 5.39 units).

Note: Unlike the previous problem, $\sqrt{29}$ is not a whole number — it is irrational. This is normal in coordinate geometry. Most distances between integer-coordinate points are irrational; only certain special pairs (like 3-4-5, 5-12-13) give nice whole-number hypotenuses.

Worked Example 3Find MA

SOLVED

Find the length of the segment from $M(9, 6)$ to $A(3, 4)$ .

Solution

Step 1 — Find the legs.

Horizontal leg = $|x_A - x_M| = |3 - 9| = 6$
Vertical leg = $|y_A - y_M| = |4 - 6| = 2$

Step 2 — Apply Baudhāyana–Pythagoras.

$MA^2 = 6^2 + 2^2 = 36 + 4 = 40$

Step 3 — Take the square root.

$MA = \sqrt{40} = 2\sqrt{10} \approx 6.32$

Answer: $MA = \sqrt{40} = 2\sqrt{10}$ units (≈ 6.32 units).

Triangle ADM has sides: $AD = 5$ , $DM = \sqrt{29} \approx 5.39$ , $MA = 2\sqrt{10} \approx 6.32$ . Notice all three are different — this is a scalene triangle, and (since no side² equals the sum of the other two squared) it is acute-angled. ✓

India's Scientific Contributions — Baudhāyana, the Original Source

Baudhāyana, who lived around 800 BCE, wrote the Śulba-sūtra — 'the rules of the cord' — as a manual for laying down sacred Vedic fire-altars. Hidden inside that practical text is the earliest known statement of the theorem we now call Pythagoras' theorem.

Practice Yourself — Distances by Pythagoras

Find the distance between each pair of points. Cover the answers and try it.

$(0, 0)$ and $(3, 4)$
$(1, 2)$ and $(4, 6)$
$(-2, 1)$ and $(1, 5)$
$(-3, -4)$ and $(0, 0)$
$(5, 12)$ and $(0, 0)$
$(2, 7)$ and $(2, 12)$
$(-1, 3)$ and $(7, 9)$
$(0, 0)$ and $(8, 15)$
$(6, 0)$ and $(0, 8)$
$(-3, 2)$ and $(3, 2)$

Answers: 1. 5 · 2. 5 · 3. 5 · 4. 5 · 5. 13 · 6. 5 (parallel to y-axis) · 7. 10 · 8. 17 · 9. 10 · 10. 6 (parallel to x-axis).

Notice: Six of these involve 3-4-5, 5-12-13, 6-8-10, or 8-15-17 Pythagorean triples — these are favourites in textbook problems because they give whole-number answers. Other pairs give $\sqrt{}$ answers, which are equally valid.

Ready to Go Beyond

On the next page the same right-triangle technique becomes a single, named formula — the distance formula — that lets you compute the distance between any two points without redrawing the triangle each time. In Class 10 you'll meet the midpoint formula and the section formula, both close cousins. In 3-D (Class 11), the formula extends naturally to $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$ — same idea, one more dimension.

End-of-Chapter Q4Plot Z and build a right-angled triangle

NCERT

Plot the point $Z(5, -6)$ on the Cartesian plane. Construct a right-angled triangle $IZN$ and find the lengths of its three sides.

Note: The choice of $I$ and $N$ is up to you — different students can choose differently. Pick any two points $I$ and $N$ such that triangle $IZN$ is right-angled at $Z$ .

Quick Check

Q1.Two points are at $(3, 0)$ and $(8, 0)$ . What is the distance between them?

AI Generation Prompt

✦Think About It

Two points sit on the same horizontal line at $(2, 5)$ and $(7, 5)$ . The distance between them is 5 units — you found this on a previous page just by subtracting the x-coordinates.

Now let me move one of those points slightly: $(2, 5)$ stays, but the other becomes $(7, 1)$ . The two points are no longer on the same line. What is the distance between them now?

Is it still 5? More? Less? And how would you compute it without a ruler?

If you draw a horizontal line from one point and a vertical line from the other, you build a right triangle. The diagonal you want is its hypotenuse.

✦बौधायन शुल्बसूत्र — १.१२

The Baudhāyana–Pythagoras Theorem

दीर्घचतुरस्रस्य अक्ष्णयारज्जुः

पार्श्वमानी तिर्यङ्मानी च

यत् पृथग् भूते कुरुतः तदुभयं करोति॥

(dīrghacaturasrasya akṣṇayārajjuḥ / pārśvamānī tiryaṅmānī ca / yat pṛthag bhūte kurutaḥ tadubhayaṃ karoti)

'The square on the diagonal of a rectangle equals the sum of the squares on the two sides.'

Easy distances first: along an axis or parallel to it

On previous pages you found distances between points whenever the segment between them was parallel to one of the axes. The rule was simple subtraction:

Two points $(x_1, y)$ and $(x_2, y)$ on the same horizontal line (same y-coordinate): the distance is $|x_2 - x_1|$ .
Two points $(x, y_1)$ and $(x, y_2)$ on the same vertical line (same x-coordinate): the distance is $|y_2 - y_1|$ .

For example, in Reiaan's room (Page 5), the room door $D_1(7.5, 0)$ to $R_1(11.5, 0)$ is horizontal — its width is $|11.5 - 7.5| = 4$ ft. Easy.

This is the moment Baudhāyana's theorem becomes a tool of coordinate geometry.

Triangle ADM: the worked example

Consider three points in Quadrant I:

$A = (3, 4)$
$D = (7, 1)$
$M = (9, 6)$

AI Generation Prompt

📸 Fig. 1.6 — Triangle ADM with $A(3, 4)$, $D(7, 1)$, $M(9, 6)$, drawn entirely inside Quadrant I.

Focus on side $AD$ first. The endpoints are $A(3, 4)$ and $D(7, 1)$ . To find $AD$ :

Drop a vertical line from $A$ down to the level of $D$ (i.e. to $y = 1$ ). Call the foot $C$ . Then $C = (3, 1)$ .
Now we have three points: $A(3, 4)$ , $C(3, 1)$ , $D(7, 1)$ . Notice that $\angle ACD = 90°$ — $C$ is directly below $A$ and directly left of $D$ , so the segment $AC$ is vertical and the segment $CD$ is horizontal. Triangle $ACD$ is right-angled at $C$ .
Now compute the two legs by simple subtraction:
- $CD$ = horizontal distance = $|7 - 3| = 4$
- $AC$ = vertical distance = $|4 - 1| = 3$
By Baudhāyana–Pythagoras: $AD^2 = AC^2 + CD^2 = 3^2 + 4^2 = 9 + 16 = 25$ , so $AD = \sqrt{25} = 5$ .

The pattern is so important it deserves a name. The distance between two points in a plane is the hypotenuse of the right triangle whose legs are parallel to the axes.

Loading simulator…

Worked Example 1Find AD

SOLVED

Find the length of the segment from $A(3, 4)$ to $D(7, 1)$ .

Solution

Step 1 — Find the legs of the right triangle.

Horizontal leg = $|x_D - x_A| = |7 - 3| = 4$
Vertical leg = $|y_A - y_D| = |4 - 1| = 3$

Step 2 — Apply Baudhāyana–Pythagoras.

$AD^2 = 4^2 + 3^2 = 16 + 9 = 25$

Step 3 — Take the square root.

$AD = \sqrt{25} = 5$

Answer: $AD = 5$ units.

Logical ReasoningLevel 3 · Analysis

Notice that for $A(3, 4)$ and $D(7, 1)$ , both simple coordinate subtractions give: $|7 - 3| = 4$ and $|4 - 1| = 3$ . Neither of these is the distance $AD = 5$ .

Why is the distance not just one of these subtractions, the way it was for points on a horizontal or vertical line?

Worked Example 2Find DM

SOLVED

Find the length of the segment from $D(7, 1)$ to $M(9, 6)$ .

Solution

Step 1 — Find the legs.

Horizontal leg = $|x_M - x_D| = |9 - 7| = 2$
Vertical leg = $|y_M - y_D| = |6 - 1| = 5$

Step 2 — Apply Baudhāyana–Pythagoras.

$DM^2 = 2^2 + 5^2 = 4 + 25 = 29$

Step 3 — Take the square root.

$DM = \sqrt{29} \approx 5.39$

Answer: $DM = \sqrt{29}$ units (≈ 5.39 units).

Worked Example 3Find MA

SOLVED

Find the length of the segment from $M(9, 6)$ to $A(3, 4)$ .

Solution

Step 1 — Find the legs.

Horizontal leg = $|x_A - x_M| = |3 - 9| = 6$
Vertical leg = $|y_A - y_M| = |4 - 6| = 2$

Step 2 — Apply Baudhāyana–Pythagoras.

$MA^2 = 6^2 + 2^2 = 36 + 4 = 40$

Step 3 — Take the square root.

$MA = \sqrt{40} = 2\sqrt{10} \approx 6.32$

Answer: $MA = \sqrt{40} = 2\sqrt{10}$ units (≈ 6.32 units).

India's Scientific Contributions — Baudhāyana, the Original Source

Ready to Go Beyond

End-of-Chapter Q4Plot Z and build a right-angled triangle

NCERT

Plot the point $Z(5, -6)$ on the Cartesian plane. Construct a right-angled triangle $IZN$ and find the lengths of its three sides.

Note: The choice of $I$ and $N$ is up to you — different students can choose differently. Pick any two points $I$ and $N$ such that triangle $IZN$ is right-angled at $Z$ .

Quick Check

Q1.Two points are at $(3, 0)$ and $(8, 0)$ . What is the distance between them?