Unit notes and practice questions

Grade 11 Mathematics unit 6: Transformations of the Plane

Everything here is for unit 6 of the Grade 11 Mathematics textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.

4
Cards: 10
Practice questions on unit 6: 313–354
The textbook pages for unit 6

What unit 6 covers

1.Translation
2.Reflection
3.Rotation
4.Applications

Short notes for unit 6

The cards filed under this chapter on the Grade 11 Mathematics page. Open one at a time.

Reflection Across the Line y = mxFormula

\begin{cases} x' = x\cos(2\theta) + y\sin(2\theta) \\ y' = x\sin(2\theta) - y\cos(2\theta) \end{cases}

\begin{cases} x' = x\cos(2\theta) + y\sin(2\theta) \\ y' = x\sin(2\theta) - y\cos(2\theta) \end{cases}

$x'$: x coordinate of the reflected point $dimensionless$
$y'$: y coordinate of the reflected point $dimensionless$
$x$: x coordinate of the original point $dimensionless$
$y$: y coordinate of the original point $dimensionless$
$\theta$: inclination angle of line y = mx where m = tan theta $\text{rad}$

When you use it

Use this formula to find the reflection of any point (x, y) across a line passing through the origin with slope m = tan(theta).

Watch out

The formula uses 2 theta inside sine and cosine, not theta.

Drafted from Grade 11 Mathematics, pages 313-354, then checked twice before it went up

Rotation About an Arbitrary PointFormula

\begin{cases} x' = a + (x - a)\cos\theta - (y - b)\sin\theta \\ y' = b + (x - a)\sin\theta + (y - b)\cos\theta \end{cases}

\begin{cases} x' = a + (x - a)\cos\theta - (y - b)\sin\theta \\ y' = b + (x - a)\sin\theta + (y - b)\cos\theta \end{cases}

$x'$: x coordinate of the image point $dimensionless$
$y'$: y coordinate of the image point $dimensionless$
$x$: x coordinate of the original point $dimensionless$
$y$: y coordinate of the original point $dimensionless$
$a$: x coordinate of the centre of rotation $dimensionless$
$b$: y coordinate of the centre of rotation $dimensionless$
$\theta$: angle of rotation $\text{rad}$

When you use it

Use this formula to find the image of a point rotated counterclockwise through an angle about any given center point (a, b).

Watch out

Do not forget to subtract the center coordinates (a, b) before applying trigonometric functions, then add them back at the end.

Drafted from Grade 11 Mathematics, pages 313-354, then checked twice before it went up

Rotations About the OriginReference table90^° · 180^° · 270^° · +1

Angle of Rotation	Image of (x, y)
$90^\circ$	$(-y, x)$
$180^\circ$	$(-x, -y)$
$270^\circ$	$(y, -x)$
$360^\circ$	$(x, y)$

When you use it

Use this table for quick evaluation of counterclockwise rotations about the origin (0, 0) by standard angles.

Watch out

Clockwise rotation by an angle corresponds to counterclockwise rotation by 360 degrees minus that angle.

Drafted from Grade 11 Mathematics, pages 313-354, then checked twice before it went up

Standard Reflections in the PlaneReference tablex axis (y = 0) · y axis (x = 0) · y = x · +1

Line of Reflection	Image of (x, y)
$x\text{ axis } (y = 0)$	$(x, -y)$
$y\text{ axis } (x = 0)$	$(-x, y)$
$y = x$	$(y, x)$
$y = -x$	$(-y, -x)$

When you use it

Use this reference table when finding the coordinates of a point reflected across common axes and diagonal lines.

Watch out

Reflecting across y = -x swaps the variables and negates both, unlike reflecting across y = x which only swaps them.

Drafted from Grade 11 Mathematics, pages 313-354, then checked twice before it went up

Practice questions on unit 6

10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.

Question 1

According to the vector summary, what is the definition of a zero vector (null vector)?

A.A vector whose magnitude is zero and whose direction is indeterminate.
B.A vector whose magnitude is one and whose direction is fixed.
C.A vector whose magnitude is undefined and whose direction is zero.
D.A vector whose magnitude is negative and direction is horizontal.

Show the answer and why

Answer: A A vector whose magnitude is zero and whose direction is indeterminate.

Why

By definition, a zero vector or null vector is a vector whose magnitude is zero and whose direction is indeterminate.

Question 2

If a circle is centred at the origin with radius r, which equation represents the tangent line to the circle at the point (x₁, y₁)?

A.x x1 + y y1 = r^2
B.x x1 - y y1 = r^2
C.(x + x1) + (y + y1) = r
D.x^2 x1 + y^2 y1 = r^2

Show the answer and why

Answer: A x x1 + y y1 = r^2

Why

When a circle is centred at the origin, the general tangent line formula simplifies directly to x x₁ + y y₁ = r².

Question 3

Transformations that preserve shape, size, and the distance between any two points are classified as which type of motion?

A.Rigid motion
B.Non-rigid motion
C.Scalar projection
D.Variable expansion

Show the answer and why

Answer: A Rigid motion

Why

A rigid motion is a transformation of the plane that preserves the shape, size, and distance between any two points.

Question 4

According to the properties of cross products, what is the cross product of any vector u with itself (u × u)?

A.0
B.|u|^2
C.1
D.-u

Show the answer and why

Answer: A 0

Why

The cross product of any vector with itself is always the zero vector, so u × u = 0.

Question 5

A student in Dire Dawa throws a ball with an initial velocity of 70 m/s at an angle of 35° horizontally. What are the horizontal and vertical components of the velocity?

A.Horizontal: 57.34 m/s, Vertical: 40.15 m/s
B.Horizontal: 40.15 m/s, Vertical: 57.34 m/s
C.Horizontal: 70.00 m/s, Vertical: 35.00 m/s
D.Horizontal: 49.50 m/s, Vertical: 49.50 m/s

Show the answer and why

Answer: A Horizontal: 57.34 m/s, Vertical: 40.15 m/s

Why

The velocity vector is v = 70 cos(35°) i + 70 sin(35°) j. Simplifying gives v = 57.34 i + 40.15 j, so the horizontal component is 57.34 m/s and the vertical component is 40.15 m/s.

Question 6

What is the equation of the tangent line through the point R(3, 4) on the circle x² + y² = 25?

A.3x + 4y = 25
B.4x + 3y = 25
C.3x - 4y = 25
D.4x - 3y = 25

Show the answer and why

Answer: A 3x + 4y = 25

Why

Using the formula x x₁ + y y₁ = r² for a circle centred at the origin with radius 5, substituting (3, 4) gives 3x + 4y = 25.

Question 7

Consider the vectors u = i + 5k and v = 10i + 3j - 2k. Are these vectors orthogonal?

A.Yes, because their dot product is 1(10) + 0(3) + 5(-2) = 0.
B.No, because their dot product is 10 + 3 - 10 = 3.
C.No, because their cross product is 0.
D.Yes, because they have the same magnitude.

Show the answer and why

Answer: A Yes, because their dot product is 1(10) + 0(3) + 5(-2) = 0.

Why

Two vectors are orthogonal if their scalar (dot) product is zero. Here, u · v = (1)(10) + (0)(3) + (5)(-2) = 10 - 10 = 0, so they are orthogonal.

Question 8

Using vector methods, the diagonals of a parallelogram meet at right angles if and only if the parallelogram is a:

A.rhombus
B.trapezoid
C.scalene quadrilateral
D.general rectangle

Show the answer and why

Answer: A rhombus

Why

The dot product of the diagonal vectors (u + v) · (v - u) = |v|² - |u|² equals 0 if and only if |v| = |u|, which means the adjacent sides are equal in length (a rhombus).

Question 9

What is the equation of the circle whose diameter has endpoints P(-1, -1) and Q(4, 11)?

A.x^2 + y^2 - 3x - 10y - 15 = 0
B.x^2 + y^2 + 3x + 10y - 15 = 0
C.x^2 + y^2 - 3x - 10y + 15 = 0
D.x^2 + y^2 - 4x - 6y - 3 = 0

Show the answer and why

Answer: A x^2 + y^2 - 3x - 10y - 15 = 0

Why

The centre is C(3/2, 5) and radius is 13/2. Expanding (x - 3/2)² + (y - 5)² = (13/2)² gives x² + y² - 3x - 10y - 15 = 0.

Question 10

Find the equation of the tangent line to the circle x² + y² + 10x + 2y + 13 = 0 at the point (-3, 2).

A.2x + 3y = 0
B.3x + 2y = 0
C.2x + 3y = 13
D.5x + y = -13

Show the answer and why

Answer: A 2x + 3y = 0

Why

Completing the square gives centre (-5, -1) and r² = 13. Using the tangent formula (x - x₀)(x₁ - x₀) + (y - y₀)(y₁ - y₀) = r² gives (x + 5)(2) + (y + 1)(3) = 13, which simplifies to 2x + 3y = 0.

The app holds 18 questions on this unit, marked as you go. These are the first 10.

The textbook pages for unit 6

Unit 6 runs from page 313 to page 354 of the Ministry of Education edition.

Ministry of Education edition, English, 494 pages

Download the PDF

All Grade 11 Mathematics short notes

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