Reflections in the Axes — Class 9 Mathematics Chapter 1

🖼 Image PendingA wide cinematic view of a mirror-still lake at dawn — a single triangular sail boat above the water, its inverted reflection perfectly mirrored below; the water surface acts as the x-axis, suggesting the geometry of reflection across an axis

AI Generation Prompt

Ultra-wide cinematic banner (16:5 ratio). A perfectly still mountain lake at dawn. A single triangular-sailed boat sits above the water on the upper half of the frame; below, the water surface acts as a mirror, returning a perfect inverted reflection of the same boat. The water line through the centre of the frame is razor-sharp — it is, in effect, the x-axis. Cool dawn light, atmospheric mist hanging over the water. The image conveys: nature performs reflections constantly; mathematics names them and writes them as coordinate transformations. Painterly cinematic illustration. Dark background. No text, no labels.

✦Think About It

When you stand in front of a mirror and lift your right hand, your reflection lifts what looks like its left hand. Lengths and angles are unchanged — the reflection is the same height, the same shape, the same proportions — but left and right have swapped.

Now take this idea to the coordinate plane. If you reflect a point across the y-axis, what happens to its coordinates? What stays the same — and what flips?

Try a specific point, say $(3, 4)$ . The y-axis is the vertical line $x = 0$ . Imagine folding the page along that line. Where does $(3, 4)$ land?

✦भगवद्गीता — २.२२ (परावर्तन का दृष्टान्त)

On Mirrored Forms

वासांसि जीर्णानि यथा विहाय

नवानि गृह्णाति नरोऽपराणि।

तथा शरीराणि विहाय जीर्णान्

अन्यानि संयाति नवानि देही॥

'जैसे मनुष्य पुराने कपड़े उतार कर नये पहन लेता है, वैसे ही आत्मा एक रूप छोड़ कर दूसरा रूप ले लेती है।'

'As a person sheds worn-out garments and puts on new ones, so the self leaves one form and takes another.'

The Gita's image of form-changing-while-essence-stays is exactly the geometry of a reflection: every coordinate of a shape may change, but lengths, angles, and the shape itself are preserved. The garment changes; what was real about it does not.

Reflecting triangle ADM in the y-axis

Recall the triangle from Page 9: $A(3, 4)$ , $D(7, 1)$ , $M(9, 6)$ — all three points in Quadrant I. Now imagine folding the coordinate plane along the y-axis. Each vertex lands at its mirror image on the other side of the y-axis. Specifically:

$A(3, 4) \to A'(-3, 4)$
$D(7, 1) \to D'(-7, 1)$
$M(9, 6) \to M'(-9, 6)$

Notice the pattern: the y-coordinate stays the same; the x-coordinate switches its sign. The image triangle $A'D'M'$ now lives in Quadrant II — a perfect mirror copy of the original.

It is natural to ask: are the three side lengths of the image triangle the same as the three side lengths of the original? Or has reflection somehow stretched, shrunk, or warped the triangle?

🖼 Image PendingCartesian plane with two triangles. The right triangle in Quadrant I has vertices A(3,4), D(7,1), M(9,6). The left triangle in Quadrant II is the y-axis-reflection: A'(-3,4), D'(-7,1), M'(-9,6). Each pair of corresponding vertices sits at the same height, on opposite sides of the y-axis. Faint dashed horizontal lines connect each pair across the y-axis. Tick marks from -9 to +9 on the x-axis and 0 to +6 on the y-axis.

AI Generation Prompt

Cartesian plane diagram on green graph paper over a dark background. The y-axis is drawn slightly thicker (highlighted) to emphasise that it is the reflection axis. In Quadrant I, three points labelled A(3,4), D(7,1), M(9,6) form a triangle (light orange). In Quadrant II, three points labelled A'(-3,4), D'(-7,1), M'(-9,6) form the mirror triangle (light green). Faint dashed horizontal lines connect each pair of corresponding points across the y-axis to emphasise the reflection. Tick marks at every integer from −9 to +9 on the x-axis and from 0 to +6 on the y-axis. Style: clean educational mathematics illustration. Dark background, orange/green accent labels, clean technical illustration style.

📸 Fig. 1.9 — Triangle ADM in Quadrant I, reflected across the y-axis to give triangle A'D'M' in Quadrant II.

Loading simulator…

Lengths are preserved — three checks

Worked Example 1A′D′ vs AD

SOLVED

Compute $A'D'$ where $A'(-3, 4)$ and $D'(-7, 1)$ , and compare it to the original $AD = 5$ .

Solution

Step 1 — Apply the distance formula.

$A'D' = \sqrt{(-7 - (-3))^2 + (1 - 4)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

Step 2 — Compare.

Original $AD = 5$ (computed on Page 9). Reflected $A'D' = 5$ . Identical.

Why? The differences inside the formula are $(-7) - (-3) = -4$ and $1 - 4 = -3$ . Compare with the original differences $7 - 3 = 4$ and $1 - 4 = -3$ . The y-difference is unchanged (since y is preserved by a y-axis reflection); the x-difference flipped sign — but the formula squares it, so the sign is irrelevant. Squaring kills the sign — that is why reflection preserves length.

Answer: $A'D' = 5$ — same as $AD$ .

Worked Example 2D′M′ vs DM

SOLVED

Compute $D'M'$ where $D'(-7, 1)$ and $M'(-9, 6)$ , and compare with the original $DM = \sqrt{29}$ .

Solution

Step 1 — Apply the distance formula.

$D'M' = \sqrt{(-9 - (-7))^2 + (6 - 1)^2} = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}$

Step 2 — Compare.

$DM = \sqrt{29}$ on Page 9. $D'M' = \sqrt{29}$ . Identical.

Answer: $D'M' = \sqrt{29}$ — same as $DM$ .

Worked Example 3M′A′ vs MA

SOLVED

Compute $M'A'$ where $M'(-9, 6)$ and $A'(-3, 4)$ , and compare with the original $MA = \sqrt{40}$ .

Solution

Step 1 — Apply the distance formula.

$M'A' = \sqrt{(-3 - (-9))^2 + (4 - 6)^2} = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40}$

Step 2 — Compare.

$MA = \sqrt{40}$ on Page 9. $M'A' = \sqrt{40}$ . Identical.

The reflected triangle has exactly the same three side-lengths as the original.

Answer: $M'A' = \sqrt{40}$ — same as $MA$ . Reflection across the y-axis preserves every side length.

The three rules — coordinate-by-coordinate

What we showed for the triangle works for any point in the plane. Let me state the three rules cleanly.

For a point $P(x, y)$ :

Reflection in the y-axis (the line $x = 0$ ): $P' = (-x, y)$ . The x-coordinate flips; y stays.
Reflection in the x-axis (the line $y = 0$ ): $P' = (x, -y)$ . The y-coordinate flips; x stays.
Reflection through the origin (point reflection): $P' = (-x, -y)$ . Both coordinates flip.

What is preserved? Every distance between every pair of points. Every angle. Every area. Every shape. (Mathematicians call any transformation with this property an isometry — iso = same, metry = measure.)

What is reversed? Orientation. A clockwise-labelled triangle becomes anti-clockwise-labelled after a single reflection — just like raising your right hand turns into your reflection raising its left.

Special case: if you reflect twice (e.g., once in the y-axis, then once in the x-axis), the orientation reverses twice — and you end up with the origin reflection, which preserves orientation. Two reflections compose into a rotation.

Spatial ReasoningLevel 4 · Evaluation

Consider triangle $ADM$ from Page 9 reflected through the origin instead of the y-axis. The new vertices are $A''(-3, -4)$ , $D''(-7, -1)$ , $M''(-9, -6)$ .

Will the side lengths of triangle $A''D''M''$ be the same as the side lengths of the original triangle $ADM$ ?

Practice Yourself — Reflections

A. Quick reflections. For each point, write its image under the three reflections (in y-axis, in x-axis, through origin):

$(2, 5)$ → ?, ?, ?
$(-3, 4)$ → ?, ?, ?
$(0, 7)$ → ?, ?, ?
$(-6, -2)$ → ?, ?, ?
$(5, 0)$ → ?, ?, ?

B. Triangle work. Triangle $PQR$ has vertices $P(2, 3)$ , $Q(5, 1)$ , $R(4, 6)$ .

Find the coordinates of the y-axis reflection $P'Q'R'$ .
Find the side lengths $PQ$ , $QR$ , $RP$ .
Find the side lengths $P'Q'$ , $Q'R'$ , $R'P'$ . Confirm they match.

C. NCERT Think and Reflect (p. 11).

What has remained the same and what has changed under reflection? Lengths, angles, areas — all unchanged. Orientation (clockwise vs anti-clockwise labelling) is reversed. Coordinates of every point are different.
Would the same observations be true if triangle $ADM$ is reflected in the x-axis instead of the y-axis? Yes — every reflection (in any axis or through any point) is an isometry; lengths and angles are always preserved.

Answers (A): 1. $(-2, 5), (2, -5), (-2, -5)$ · 2. $(3, 4), (-3, -4), (3, -4)$ · 3. $(0, 7), (0, -7), (0, -7)$ · 4. $(6, -2), (-6, 2), (6, 2)$ · 5. $(-5, 0), (5, 0), (-5, 0)$ . (B) 6. $P'(-2, 3), Q'(-5, 1), R'(-4, 6)$ . 7. $PQ = \sqrt{13}$ , $QR = \sqrt{26}$ , $RP = \sqrt{13}$ . 8. Identical to (7) — reflection preserves all three.

Ready to Go Beyond

Reflections are part of a much larger family of plane transformations you'll meet in higher mathematics:

Quick Check

Q1.What is the image of $(7, -2)$ under reflection in the x-axis?

AI Generation Prompt

✦Think About It

Now take this idea to the coordinate plane. If you reflect a point across the y-axis, what happens to its coordinates? What stays the same — and what flips?

Try a specific point, say $(3, 4)$ . The y-axis is the vertical line $x = 0$ . Imagine folding the page along that line. Where does $(3, 4)$ land?

✦भगवद्गीता — २.२२ (परावर्तन का दृष्टान्त)

On Mirrored Forms

वासांसि जीर्णानि यथा विहाय

नवानि गृह्णाति नरोऽपराणि।

तथा शरीराणि विहाय जीर्णान्

अन्यानि संयाति नवानि देही॥

'As a person sheds worn-out garments and puts on new ones, so the self leaves one form and takes another.'

Reflecting triangle ADM in the y-axis

$A(3, 4) \to A'(-3, 4)$
$D(7, 1) \to D'(-7, 1)$
$M(9, 6) \to M'(-9, 6)$

Notice the pattern: the y-coordinate stays the same; the x-coordinate switches its sign. The image triangle $A'D'M'$ now lives in Quadrant II — a perfect mirror copy of the original.

It is natural to ask: are the three side lengths of the image triangle the same as the three side lengths of the original? Or has reflection somehow stretched, shrunk, or warped the triangle?

AI Generation Prompt

📸 Fig. 1.9 — Triangle ADM in Quadrant I, reflected across the y-axis to give triangle A'D'M' in Quadrant II.

Loading simulator…

Lengths are preserved — three checks

Worked Example 1A′D′ vs AD

SOLVED

Compute $A'D'$ where $A'(-3, 4)$ and $D'(-7, 1)$ , and compare it to the original $AD = 5$ .

Solution

Step 1 — Apply the distance formula.

$A'D' = \sqrt{(-7 - (-3))^2 + (1 - 4)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

Step 2 — Compare.

Original $AD = 5$ (computed on Page 9). Reflected $A'D' = 5$ . Identical.

Answer: $A'D' = 5$ — same as $AD$ .

Worked Example 2D′M′ vs DM

SOLVED

Compute $D'M'$ where $D'(-7, 1)$ and $M'(-9, 6)$ , and compare with the original $DM = \sqrt{29}$ .

Solution

Step 1 — Apply the distance formula.

$D'M' = \sqrt{(-9 - (-7))^2 + (6 - 1)^2} = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}$

Step 2 — Compare.

$DM = \sqrt{29}$ on Page 9. $D'M' = \sqrt{29}$ . Identical.

Answer: $D'M' = \sqrt{29}$ — same as $DM$ .

Worked Example 3M′A′ vs MA

SOLVED

Compute $M'A'$ where $M'(-9, 6)$ and $A'(-3, 4)$ , and compare with the original $MA = \sqrt{40}$ .

Solution

Step 1 — Apply the distance formula.

$M'A' = \sqrt{(-3 - (-9))^2 + (4 - 6)^2} = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40}$

Step 2 — Compare.

$MA = \sqrt{40}$ on Page 9. $M'A' = \sqrt{40}$ . Identical.

The reflected triangle has exactly the same three side-lengths as the original.

Answer: $M'A' = \sqrt{40}$ — same as $MA$ . Reflection across the y-axis preserves every side length.

The three rules — coordinate-by-coordinate

What we showed for the triangle works for any point in the plane. Let me state the three rules cleanly.

For a point $P(x, y)$ :

Reflection in the y-axis (the line $x = 0$ ): $P' = (-x, y)$ . The x-coordinate flips; y stays.
Reflection in the x-axis (the line $y = 0$ ): $P' = (x, -y)$ . The y-coordinate flips; x stays.
Reflection through the origin (point reflection): $P' = (-x, -y)$ . Both coordinates flip.

Spatial ReasoningLevel 4 · Evaluation

Consider triangle $ADM$ from Page 9 reflected through the origin instead of the y-axis. The new vertices are $A''(-3, -4)$ , $D''(-7, -1)$ , $M''(-9, -6)$ .

Will the side lengths of triangle $A''D''M''$ be the same as the side lengths of the original triangle $ADM$ ?

Ready to Go Beyond

Reflections are part of a much larger family of plane transformations you'll meet in higher mathematics:

Quick Check

Q1.What is the image of $(7, -2)$ under reflection in the x-axis?