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Shape and Space - Transformations (Translation, Reflection, Rotation)

Grade 6IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Translation involves sliding a shape to a new position without rotating it or changing its size. Every point moves the same distance in the same direction, defined by a vector (xy)\binom{x}{y}, where xx is the horizontal shift and yy is the vertical shift.

A triangle translated by vector (2, 1) on a coordinate plane.
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Reflection creates a mirror image of a shape across a 'line of reflection'. For Grade 6, we focus on reflections across the xx-axis and yy-axis. Every point and its image are equidistant from this line.

A rectangle reflected across the y-axis.
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Rotation turns a figure about a fixed point called the center of rotation. We describe the turn using an angle (e.g., 90∘90^{\circ} or 180∘180^{\circ}) and a direction (Clockwise or Counter-clockwise).

A 90 degree rotation arc centered at the origin.
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Invariant properties: In translation, reflection, and rotation, the size and shape of the object remain the same; these are called 'isometries'. Only the position or orientation changes.

📐Formulae

Translation by vector (ab):(x,y)→(x+a,y+b)\binom{a}{b}: (x, y) \rightarrow (x + a, y + b)

Reflection across the xx-axis: (x,y)→(x,−y)(x, y) \rightarrow (x, -y)

Reflection across the yy-axis: (x,y)→(−x,y)(x, y) \rightarrow (-x, y)

Rotation of 180∘180^{\circ} about the origin: (x,y)→(−x,−y)(x, y) \rightarrow (-x, -y)

Rotation of 90∘90^{\circ} Clockwise about the origin: (x,y)→(y,−x)(x, y) \rightarrow (y, -x)

Rotation of 90∘90^{\circ} Counter-clockwise about the origin: (x,y)→(−y,x)(x, y) \rightarrow (-y, x)

💡Examples

Problem 1:

A triangle has vertices at A(1,2)A(1, 2), B(4,2)B(4, 2), and C(1,5)C(1, 5). Translate the triangle using the vector (−32)\binom{-3}{2} and state the new coordinates of vertex A′A'.

Solution:

  1. Identify the translation values: a=−3a = -3 (move 3 units left) and b=2b = 2 (move 2 units up).
  2. Apply the formula (x+a,y+b)(x + a, y + b) to the original coordinates of A(1,2)A(1, 2).
  3. Calculate the new x-coordinate: x′=1+(−3)=−2x' = 1 + (-3) = -2.
  4. Calculate the new y-coordinate: y′=2+2=4y' = 2 + 2 = 4.
  5. The new coordinates for vertex A′A' are (−2,4)(-2, 4).

Explanation:

To translate a point, we add the horizontal component of the vector to the xx coordinate and the vertical component to the yy coordinate.

Problem 2:

Point PP is located at (3,−5)(3, -5). Reflect point PP across the yy-axis and then describe the position of the resulting image P′P'.

Solution:

  1. Identify the reflection rule for the yy-axis: (x,y)→(−x,y)(x, y) \rightarrow (-x, y).
  2. The original xx coordinate is 33, so the new xx coordinate is −(3)=−3-(3) = -3.
  3. The yy coordinate remains unchanged: y=−5y = -5.
  4. The image P′P' is located at (−3,−5)(-3, -5).

Explanation:

When reflecting across the yy-axis, the point moves horizontally to the opposite side of the vertical axis, so only the sign of the xx-coordinate changes.

Problem 3:

A square has vertices at S(1,−1)S(1, -1), T(3,−1)T(3, -1), U(3,−3)U(3, -3), and V(1,−3)V(1, -3). Reflect the square across the xx-axis and state the new coordinates of vertex U′U'.

Square STUV reflected across the x-axis to form square S'T'U'V'.

Solution:

  1. Identify the rule for reflection across the xx-axis: (x,y)→(x,−y)(x, y) \rightarrow (x, -y).
  2. Apply the rule to point U(3,−3)U(3, -3).
  3. New xx-coordinate remains 33.
  4. New yy-coordinate becomes −(−3)=3-(-3) = 3.
  5. Therefore, U′U' is at (3,3)(3, 3).

Explanation:

Reflecting across the xx-axis flips the shape vertically. The horizontal distance from the yy-axis stays the same, but the sign of the yy-coordinate is inverted.

Problem 4:

Triangle ABCABC has vertices A(0,0)A(0, 0), B(2,0)B(2, 0), and C(0,3)C(0, 3). Rotate the triangle 90∘90^{\circ} Clockwise about the origin and find the new coordinates of C′C'.

Right triangle rotated 90 degrees clockwise about the origin.

Solution:

  1. Identify the rule for 90∘90^{\circ} Clockwise rotation: (x,y)→(y,−x)(x, y) \rightarrow (y, -x).
  2. Apply the rule to vertex C(0,3)C(0, 3).
  3. The new xx-coordinate is the old yy-coordinate: 33.
  4. The new yy-coordinate is the negative of the old xx-coordinate: −(0)=0-(0) = 0.
  5. Thus, C′C' is (3,0)(3, 0).

Explanation:

In a 90∘90^{\circ} clockwise rotation, the vertical component of the original point becomes the horizontal component of the new point, and the horizontal component is flipped to the negative yy-direction.