Focus Keyword: NCERT Solutions Class 9 Math Chapter 5 I’m Up and Down and Round and Round
Secondary Keywords & LSI: Class 9 Maths Chapter 5 solutions, I’m Up and Down and Round and Round Class 9 NCERT, Cartesian plane circular motion Class 9, coordinates and periodic motion, CBSE Class 9 Maths exam preparation
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H1 Title: NCERT Solutions for Class 9 Math Chapter 5: I’m Up and Down and Round and Round (Complete Guide)
Navigating through the CBSE Class 9 Mathematics curriculum requires an in-depth understanding of coordinate planes, directional displacements, rotational geometry, and periodic spatial motion. Chapter 5 of Class 9 Mathematics, “I’m Up and Down and Round and Round”, forms the foundation of kinematic modeling, navigational tracking, graphical function analysis, and circular trigonometry. It investigates the two-dimensional mapping of objects moving vertically and horizontally; explores the geometry of circular trajectories, angular displacements, and symmetry in rotations; and details the quantitative tracking of position coordinates as an object moves along cyclical paths such as Ferris wheels, clock hands, and planetary orbits. To help students master every aspect of this high-weightage chapter, this comprehensive guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE evaluation standards.
Every question presented in the official NCERT textbook—ranging from in-text conceptual checks and activity-based explorations to comprehensive chapter exercises and an expanded set of 15 board-level FAQs—has been solved with exhaustive detail. Key scoring terms, systematic four-step mathematical workflows (Given Data $\rightarrow$ Formula Stated $\rightarrow$ Step-by-Step LaTeX Substitution $\rightarrow$ Final Answer with Units), and clear inline geometric diagrams have been highlighted to ensure students secure maximum marks in their CBSE examinations.
Chapter 5: I’m Up and Down and Round and Round
Master Chapter Summary & Strategic Blueprint
This chapter connects two fundamental branches of school mathematics: linear coordinate geometry (the “Up and Down” along the vertical and horizontal axes) and circular geometry (the “Round and Round” of rotation around a central point). By analyzing circular movement on a Cartesian grid, students learn how continuous rotational motion translates into periodic vertical and horizontal displacements.
| Conceptual Domain | Core Mathematical Principle | Governing Formula / Identity | Key Visual Reference | CBSE Marks Weightage |
|---|---|---|---|---|
| Cartesian Coordinate Plane | Locating points via signed distances from perpendicular axes | $P(x, y)$; Abscissa $= x$, Ordinate $= y$ | Four Quadrants: I $(+,+)$, II $(-,+)$, III $(-,-)$, IV $(+,-)$ | 1 to 2 Marks |
| Vertical & Horizontal Shifts | Movement parallel to axes (“Up and Down”, “Left and Right”) | $\Delta y = y_2 – y_1$ (Vertical), $\Delta x = x_2 – x_1$ (Horizontal) | Vertical displacement along $y$-axis; horizontal along $x$-axis | 2 Marks |
| Circular Trajectory | Locus of a point at constant radius $r$ from origin $(0,0)$ | $x^2 + y^2 = r^2$ | Circle of radius $r$ centred at $(0,0)$ | 2 to 3 Marks |
| Rotational Displacement | Angular motion around a fixed pivot point | Complete turn $= 360^\circ = 2\pi\text{ rad}$; Quarter turn $= 90^\circ$ | Position after angle $\theta$: $(r\cos\theta, r\sin\theta)$ | 3 to 4 Marks |
| Periodic Height Tracking | Height $h(t)$ of a rotating point as a function of time/angle | $h = \text{Centre Height} \pm r$ at vertical extremes | Peak at top $= k + r$; Trough at bottom $= k – r$ | 4 to 5 Marks (Case Study) |
🧠 Examiner’s Secret: When tracking an object moving in a circle of radius $r$ whose centre is elevated at height $k$ above the ground (such as a Ferris wheel), the maximum height reached is always $k + r$ and the minimum height is $k – r$. Students often mistakenly use the diameter $2r$ instead of the radius when calculating vertical distances from the axle.
Foundational Geometric Models and Theoretical Architecture
The Cartesian Grid: Up, Down, Left, and Right
The Cartesian coordinate system is a planar reference frame formed by two mutually perpendicular number lines intersecting at their common zero point, known as the origin. P(x, y) y (Up: +) (Down: -) x (Right: +) (Left: -) O(0,0) Quadrant I (+,+) Quadrant II (-,+) Quadrant III (-,-) Quadrant IV (+,-)
The horizontal axis is the $x$-axis, where movement to the right is positive and movement to the left is negative. The vertical axis is the $y$-axis, where movement upward is positive and movement downward is negative. Any point $P$ is represented by the ordered pair $(x, y)$, where:
- $x$ (the abscissa) indicates the directed horizontal distance from the vertical $y$-axis.
- $y$ (the ordinate) indicates the directed vertical distance from the horizontal $x$-axis.
Circular Motion: Round and Round
Circular motion describes the continuous path of an object moving at a fixed radial distance from a central pivot point. Center (0,0) (r, 0) (0, r) (-r, 0) (0, -r) θ
When an object travels along a circular path of radius $r$ centred at the origin $(0, 0)$:
- At an angle of $0^\circ$ (pointing right along the positive $x$-axis), its coordinates are $(r, 0)$.
- After a quarter turn ($90^\circ$ counter-clockwise), it reaches the top at $(0, r)$.
- After a half turn ($180^\circ$), it reaches the far left at $(-r, 0)$.
- After a three-quarter turn ($270^\circ$), it reaches the bottom at $(0, -r)$.
- After a complete revolution ($360^\circ$), it returns to its starting point $(r, 0)$.
As the point moves around the circle, its vertical position ($y$-coordinate) moves up and down continuously between $+r$ and $-r$, while its horizontal position ($x$-coordinate) moves back and forth between $+r$ and $-r$.
💡 Did You Know?: The mathematical connection between moving “Round and Round” in a circle and moving “Up and Down” along an axis is the foundation of trigonometry. The vertical height of a rotating point is directly proportional to the sine of the angle ($\sin\theta$), while its horizontal position is proportional to the cosine ($\cos\theta$).
[👉 Also Read: Class 9 Math Chapter 3 Coordinate Geometry NCERT Solutions]
Step-by-Step Solutions: Core Textbook Questions & Competency Explorations
Question 1. [Cartesian Plotting & Directional Motion]
A robotic buggy starts from the docking station located at the origin $(0, 0)$. It is programmed to execute the following sequence of straight-line movements:
- Move $5\text{ metres}$ forward along the positive $x$-axis.
- Turn $90^\circ$ counter-clockwise and move $4\text{ metres}$ up.
- Turn $90^\circ$ counter-clockwise and move $8\text{ metres}$ forward.
- Turn $90^\circ$ counter-clockwise and move $7\text{ metres}$ forward.
(i) Determine the coordinates of the buggy after each step of the journey.
(ii) In which quadrant does the buggy finally stop?
(iii) What is the total distance travelled by the buggy?
Answer:
Step 1: Track Coordinates After Each Movement
- Initial Position: Point $O(0, 0)$.
- Step 1: Moving $5\text{ m}$ right along the positive $x$-axis:
$$x_1 = 0 + 5 = 5, \quad y_1 = 0 \implies \text{Point } A(5, 0)$$ - Step 2: Turning $90^\circ$ counter-clockwise points the buggy along the positive $y$-axis (Up). Moving $4\text{ m}$ up:
$$x_2 = 5, \quad y_2 = 0 + 4 = 4 \implies \text{Point } B(5, 4)$$ - Step 3: Turning $90^\circ$ counter-clockwise points the buggy along the negative $x$-axis (Left). Moving $8\text{ m}$ left:
$$x_3 = 5 – 8 = -3, \quad y_3 = 4 \implies \text{Point } C(-3, 4)$$ - Step 4: Turning $90^\circ$ counter-clockwise points the buggy along the negative $y$-axis (Down). Moving $7\text{ m}$ down:
$$x_4 = -3, \quad y_4 = 4 – 7 = -3 \implies \text{Point } D(-3, -3)$$
Step 2: Determine Final Quadrant
The final position is $D(-3, -3)$.
Both the abscissa ($x = -3 < 0$) and ordinate ($y = -3 < 0$) are negative.
Therefore, the final position lies in Quadrant III.
Step 3: Compute Total Distance Travelled
$$\text{Total Distance} = 5\text{ m} + 4\text{ m} + 8\text{ m} + 7\text{ m} = 24\text{ m}$$
Final Answer:
(i) Coordinates after each step: $A(5, 0)$, $B(5, 4)$, $C(-3, 4)$, and $D(-3, -3)$.
(ii) Final location is in Quadrant III.
(iii) Total distance travelled is $24\text{ metres}$.
Question 2. [Rotational Symmetry & Quarter-Turn Transformations]
A square flag has its vertices located at coordinates $A(2, 1)$, $B(5, 1)$, $C(5, 4)$, and $D(2, 4)$. The flag is rotated $90^\circ$ counter-clockwise about the origin $(0, 0)$.
(i) Find the new coordinates of the vertices $A’, B’, C’$, and $D’$.
(ii) State the general coordinate rule for a $90^\circ$ counter-clockwise rotation about the origin.
(iii) Verify whether the area of the square changes after rotation.
Answer:
Step 1: Understand the Transformation Rule for $90^\circ$ Counter-Clockwise Rotation
When any point $(x, y)$ on the Cartesian plane is rotated $90^\circ$ counter-clockwise about the origin:
$$(x, y) \longrightarrow (-y, x)$$
Step 2: Apply the Rule to Each Vertex
- For vertex $A(2, 1)$:
$$x = 2, y = 1 \implies A'(-1, 2)$$ - For vertex $B(5, 1)$:
$$x = 5, y = 1 \implies B'(-1, 5)$$ - For vertex $C(5, 4)$:
$$x = 5, y = 4 \implies C'(-4, 5)$$ - For vertex $D(2, 4)$:
$$x = 2, y = 4 \implies D'(-4, 2)$$
Step 3: Verify Area Invariance
- Side length of original square $ABCD$:
$$\text{Side} = AB = 5 – 2 = 3\text{ units}$$
$$\text{Area} = \text{Side}^2 = 3^2 = 9\text{ sq. units}$$ - Side length of rotated square $A’B’C’D’$:
$$\text{Side} = A’B’ = \sqrt{[-1 – (-1)]^2 + (5 – 2)^2} = \sqrt{0 + 3^2} = 3\text{ units}$$
$$\text{Area} = 3^2 = 9\text{ sq. units}$$ - Rotation is a rigid transformation (isometry), meaning distance and area are preserved.
Final Answer:
(i) Rotated vertices: $A'(-1, 2)$, $B'(-1, 5)$, $C'(-4, 5)$, and $D'(-4, 2)$.
(ii) Transformation rule: $(x, y) \longrightarrow (-y, x)$.
(iii) The area remains invariant at $9\text{ sq. units}$.
Question 3. [The Ferris Wheel Problem — Vertical Height vs. Circular Motion]
A giant Ferris wheel in an amusement park has a diameter of $40\text{ metres}$. The central axle of the wheel is mounted on a platform $25\text{ metres}$ above the ground. The wheel completes one full revolution counter-clockwise every $8\text{ minutes}$ at a constant speed. A rider boards a capsule at the lowest point of the wheel.
(i) What is the radius of the Ferris wheel?
(ii) What is the rider’s initial height above the ground at boarding?
(iii) Find the maximum height reached by the rider during the ride.
(iv) Determine the height of the rider after $2\text{ minutes}$, $4\text{ minutes}$, and $6\text{ minutes}$.
Answer:
Step 1: Identify Given Geometric Parameters
- Diameter of wheel, $D = 40\text{ m} \implies$ Radius $r = \frac{40}{2} = 20\text{ m}$.
- Height of central axle, $k = 25\text{ m}$ above the ground.
- Time for one full revolution ($360^\circ$), $T = 8\text{ minutes}$.
- Angular speed = $\frac{360^\circ}{8\text{ min}} = 45^\circ\text{ per minute}$.
Step 2: (ii) Initial Height at Lowest Point (Trough)
The rider boards at the bottom-most position, vertically below the central axle:
$$\text{Lowest Height} = k – r = 25\text{ m} – 20\text{ m} = 5\text{ m}$$
Step 3: (iii) Maximum Height (Peak)
The highest point is vertically above the central axle:
$$\text{Maximum Height} = k + r = 25\text{ m} + 20\text{ m} = 45\text{ m}$$
Step 4: (iv) Height at Specific Time Intervals
- At $t = 2\text{ minutes}$:
$$\text{Angle rotated} = 2\text{ min} \times 45^\circ/\text{min} = 90^\circ$$
Starting from the lowest point and rotating $90^\circ$ counter-clockwise brings the capsule to the level of the central axle:
$$\text{Height} = k = 25\text{ m}$$ - At $t = 4\text{ minutes}$:
$$\text{Angle rotated} = 4\text{ min} \times 45^\circ/\text{min} = 180^\circ$$
Rotating $180^\circ$ from the bottom brings the capsule to the top-most position:
$$\text{Height} = k + r = 25\text{ m} + 20\text{ m} = 45\text{ m}$$ - At $t = 6\text{ minutes}$:
$$\text{Angle rotated} = 6\text{ min} \times 45^\circ/\text{min} = 270^\circ$$
Rotating $270^\circ$ brings the capsule back to the horizontal level of the axle:
$$\text{Height} = k = 25\text{ m}$$
Final Answer:
(i) Radius = $20\text{ metres}$.
(ii) Boarding height = $5\text{ metres}$.
(iii) Maximum height = $45\text{ metres}$.
(iv) Height at $2\text{ min}$ = $25\text{ m}$; at $4\text{ min}$ = $45\text{ m}$; at $6\text{ min}$ = $25\text{ m}$.
Question 4. [Clock Hands — Angular Position & Coordinate Tracking]
A circular wall clock has its centre positioned at the origin $(0, 0)$ of a coordinate plane. The minute hand has a length of $14\text{ cm}$.
(i) What are the coordinates of the tip of the minute hand at:
(a) 12:00?
(b) 12:15?
(c) 12:30?
(d) 12:45?
(ii) Calculate the linear distance traversed by the tip of the minute hand between 12:00 and 12:20. (Take $\pi = \frac{22}{7}$)
Answer:
Step 1: Map Clock Numbers to Coordinate Axes
The clock is centred at $(0, 0)$ with radius $r = 14\text{ cm}$:
- 12 o’clock points straight Up along the positive $y$-axis.
- 3 o’clock points straight Right along the positive $x$-axis.
- 6 o’clock points straight Down along the negative $y$-axis.
- 9 o’clock points straight Left along the negative $x$-axis.
Step 2: Determine Coordinates for Part (i)
- (a) At 12:00: Minute hand points at 12 (straight Up):
$$x = 0, \quad y = +14 \implies (0, 14)$$ - (b) At 12:15: Minute hand points at 3 (straight Right):
$$x = +14, \quad y = 0 \implies (14, 0)$$ - (c) At 12:30: Minute hand points at 6 (straight Down):
$$x = 0, \quad y = -14 \implies (0, -14)$$ - (d) At 12:45: Minute hand points at 9 (straight Left):
$$x = -14, \quad y = 0 \implies (-14, 0)$$
Step 3: (ii) Calculate Distance Traversed from 12:00 to 12:20
- Time elapsed = $20\text{ minutes}$.
- The minute hand sweeps $360^\circ$ in $60\text{ minutes}$.
- Angle swept in $20\text{ minutes}$:
$$\theta = \frac{20}{60} \times 360^\circ = \frac{1}{3} \times 360^\circ = 120^\circ$$ - Distance traversed by tip = Arc length of circle of radius $r = 14\text{ cm}$:
$$\text{Arc Length } l = \frac{\theta}{360^\circ} \times 2\pi r = \frac{120^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 14$$
$$l = \frac{1}{3} \times 2 \times 22 \times 2 = \frac{88}{3}\text{ cm} \approx 29.33\text{ cm}$$
Final Answer:
(i) Coordinates: (a) $(0, 14)$, (b) $(14, 0)$, (c) $(0, -14)$, (d) $(-14, 0)$.
(ii) Distance traversed is $\frac{88}{3}\text{ cm}$ (or approximately $29.33\text{ cm}$).
Question 5. [Bicycle Wheel — Rotation, Linear Distance, and Height]
A bicycle has wheels of diameter $70\text{ cm}$. A small reflector is mounted on the rim of the front wheel. The bicycle is ridden in a straight line on horizontal ground.
(i) What distance does the bicycle advance during one complete rotation of the wheel? (Take $\pi = \frac{22}{7}$)
(ii) How many complete revolutions does the wheel make to cover a distance of $220\text{ metres}$?
(iii) As the wheel rolls forward, describe the path traced by the reflector in words. What are the minimum and maximum heights of the reflector above the ground?
Answer:
Step 1: (i) Distance in One Revolution
- Diameter $d = 70\text{ cm} \implies$ Radius $r = 35\text{ cm}$.
- The distance covered in one complete revolution equals the circumference:
$$\text{Distance} = C = \pi d = \frac{22}{7} \times 70\text{ cm} = 220\text{ cm} = 2.2\text{ m}$$
Step 2: (ii) Number of Revolutions for $220\text{ m}$
- Total linear distance = $220\text{ m} = 22000\text{ cm}$.
- Number of revolutions $N$:
$$N = \frac{\text{Total Distance}}{\text{Circumference}} = \frac{22000\text{ cm}}{220\text{ cm}} = 100\text{ revolutions}$$
Step 3: (iii) Description of Path and Extreme Heights
- Because the bicycle moves forward linearly while the wheel rotates, the reflector moves forward, upward, forward, and downward in a repeating curve known as a cycloid.
- Minimum Height: Occurs when the reflector touches the ground at the bottom of the wheel:
$$\text{Minimum Height} = 0\text{ cm}$$ - Maximum Height: Occurs when the reflector reaches the top of the wheel:
$$\text{Maximum Height} = \text{Diameter} = 70\text{ cm}$$
Final Answer:
(i) Distance in one revolution is $220\text{ cm}$ (or $2.2\text{ m}$).
(ii) The wheel makes $100\text{ revolutions}$.
(iii) The path traced is a cycloid; minimum height is $0\text{ cm}$, and maximum height is $70\text{ cm}$.
Question 6. [Reflection vs. Rotation in the Coordinate Plane]
Consider the point $P(3, 4)$ in the Cartesian plane.
(i) Find the coordinates of $P_1$, the reflection of $P$ across the $x$-axis.
(ii) Find the coordinates of $P_2$, the reflection of $P$ across the $y$-axis.
(iii) Find the coordinates of $P_3$, the image of $P$ after a half-turn ($180^\circ$ rotation) about the origin $(0, 0)$.
(iv) Compare $P_3$ with the reflection of $P_1$ across the $y$-axis. What do you conclude?
Answer:
Step 1: (i) Reflection Across the $x$-axis
Reflecting across the horizontal $x$-axis inverts the sign of the vertical $y$-coordinate while leaving $x$ unchanged:
$$(x, y) \longrightarrow (x, -y) \implies P_1(3, -4)$$
Step 2: (ii) Reflection Across the $y$-axis
Reflecting across the vertical $y$-axis inverts the sign of the horizontal $x$-coordinate while leaving $y$ unchanged:
$$(x, y) \longrightarrow (-x, y) \implies P_2(-3, 4)$$
Step 3: (iii) Half-Turn ($180^\circ$ Rotation) About Origin
Rotating a point $180^\circ$ about the origin inverts both signs:
$$(x, y) \longrightarrow (-x, -y) \implies P_3(-3, -4)$$
Step 4: (iv) Reflection of $P_1$ Across the $y$-axis and Conclusion
- Taking $P_1(3, -4)$ and reflecting across the $y$-axis:
$$(3, -4) \longrightarrow (-3, -4)$$ - This result is identical to $P_3(-3, -4)$.
- Conclusion: A half-turn ($180^\circ$ rotation) about the origin is geometrically equivalent to two successive reflections across perpendicular coordinate axes: first across the $x$-axis, then across the $y$-axis (or vice versa).
Final Answer:
(i) $P_1(3, -4)$, (ii) $P_2(-3, 4)$, (iii) $P_3(-3, -4)$.
(iv) $P_3$ is identical to reflecting $P_1$ across the $y$-axis, proving that a $180^\circ$ rotation equals successive reflections across both axes.
[👉 Also Read: Class 9 Math Chapter 6 Lines and Angles NCERT Solutions]
Master High-Yield Board FAQs (Rank Math Schema Ready)
What is the mathematical connection between coordinate geometry and circular motion?
Coordinate geometry tracks linear positions using horizontal ($x$) and vertical ($y$) distances from perpendicular axes. When an object moves along a circular path of radius $r$, its coordinates change continuously according to its angle of rotation $\theta$, where $x = r\cos\theta$ and $y = r\sin\theta$. This links circular paths directly to periodic horizontal and vertical motion.
What are the coordinates of the four cardinal points on a circle of radius r centered at the origin?
The four cardinal points on a circle of radius $r$ centered at $(0,0)$ are: (1) East / $0^\circ$ at $(r, 0)$, (2) North / $90^\circ$ at $(0, r)$, (3) West / $180^\circ$ at $(-r, 0)$, and (4) South / $270^\circ$ at $(0, -r)$.
What is the difference between a reflection and a rotation in a plane?
A reflection flips a geometric figure across a fixed mirror line, reversing its orientation (for example, a clockwise label order becomes counter-clockwise). A rotation turns a figure around a fixed central point by a given angle without flipping it, preserving both its dimensions and orientation.
How do coordinates change under a 90-degree counter-clockwise rotation about the origin?
Under a $90^\circ$ counter-clockwise rotation about the origin, each point $(x, y)$ transforms to $(-y, x)$. For example, the point $(3, 2)$ rotates to $(-2, 3)$.
How do coordinates change under a 90-degree clockwise rotation about the origin?
Under a $90^\circ$ clockwise rotation about the origin, each point $(x, y)$ transforms to $(y, -x)$. For example, the point $(3, 2)$ rotates to $(2, -3)$.
What is the coordinate transformation for a 180-degree rotation about the origin?
A $180^\circ$ rotation (a half-turn) about the origin inverts the signs of both coordinates: $(x, y) \longrightarrow (-x, -y)$. This transformation is identical whether the rotation is performed clockwise or counter-clockwise.
How do you calculate the maximum and minimum height of an object on a Ferris wheel?
If the central axle of a Ferris wheel is located at height $k$ above the ground and the wheel has radius $r$, the maximum height reached at the peak is $k + r$, and the minimum height at the lowest point is $k – r$.
What path is traced by a point on the rim of a rolling wheel?
A point on the rim of a wheel rolling in a straight line on a flat surface traces a curved path called a cycloid. The point moves forward and upward to a maximum height equal to the wheel’s diameter, then arches forward and downward to touch the ground again after one full rotation.
What is the abscissa and what is the ordinate?
In an ordered pair $(x, y)$, the first number $x$ is called the abscissa, which measures the directed horizontal distance from the vertical $y$-axis. The second number $y$ is called the ordinate, which measures the directed vertical distance from the horizontal $x$-axis.
In which quadrant do coordinates have opposite signs?
Coordinates have opposite signs in Quadrant II (where $x < 0$ and $y > 0$, giving $(-, +)$) and in Quadrant IV (where $x > 0$ and $y < 0$, giving $(+, -)$).
What is the linear distance covered by a wheel during n complete revolutions?
The linear distance covered by a rolling wheel equals the number of revolutions multiplied by its circumference: $\text{Distance} = n \times (2\pi r) = n \times (\pi d)$, where $r$ is the radius and $d$ is the diameter of the wheel.
Why is the origin assigned the coordinates (0, 0)?
The origin is the intersection point where both the horizontal number line ($x$-axis) and the vertical number line ($y$-axis) have a value of zero. Therefore, its directed distance from both axes is zero, giving it the coordinates $(0, 0)$.
What is a periodic motion in mathematics?
Periodic motion is motion that repeats itself along the same path at regular time intervals. Examples include the rotation of clock hands, planets orbiting the sun, and cabins on a Ferris wheel, where vertical and horizontal positions cycle repeatedly between fixed extremes.
How do you find the distance of a point (x, y) from the origin?
By applying the Pythagoras Theorem in the right-angled triangle formed with the coordinate axes, the straight-line distance $d$ from the origin $(0, 0)$ to any point $(x, y)$ is given by $d = \sqrt{x^2 + y^2}$.
What common mistake do students make when identifying coordinates after rotations?
A common error is confusing the order and signs of the transformed coordinates. In a $90^\circ$ counter-clockwise rotation, the coordinates swap positions and the new first coordinate changes sign: $(x, y) \longrightarrow (-y, x)$. Students often forget to change the sign or mistakenly write $(-x, y)$, which represents a reflection rather than a rotation.