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Geometry and Measurement - Geometric Transformations (Translation, Reflection, Rotation)

Grade 7IB

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

🔑Concepts

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Geometric transformations describe how a shape moves or changes on a coordinate plane. Translation is a 'slide' where every point of a figure moves the same distance in the same direction. It is often represented by a column vector (ab)\begin{pmatrix} a \\ b \end{pmatrix}, where aa is the horizontal shift and bb is the vertical shift.

A triangle translated by the vector (3, 1) on a coordinate plane.
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Reflection is a 'flip' over a line called the line of reflection. Each point and its image are equidistant from this line. For example, reflecting a point (x,y)(x, y) across the xx-axis changes its coordinates to (x,−y)(x, -y).

A triangle reflected across the x-axis.
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Rotation is a 'turn' around a fixed point called the center of rotation. A rotation is defined by the center, the angle of rotation (e.g., 90∘90^\circ or 180∘180^\circ), and the direction (clockwise or counter-clockwise).

A triangle rotated 90 degrees counter-clockwise about the origin.
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Transformation Invariance: Under translation, reflection, and rotation, the original shape (pre-image) and the final shape (image) are congruent. This means their side lengths and interior angles remain identical; only their position or orientation changes.

📐Formulae

Translation Rule: (x,y)→(x+a,y+b)(x, y) \rightarrow (x + a, y + b) for a vector (ab)\begin{pmatrix} a \\ b \end{pmatrix}

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

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

Reflection in the line y=xy = x: (x,y)→(y,x)(x, y) \rightarrow (y, x)

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

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

Rotation 270∘270^\circ counter-clockwise (or 90∘90^\circ clockwise) about the origin: (x,y)→(y,−x)(x, y) \rightarrow (y, -x)

💡Examples

Problem 1:

A triangle has vertices A(1,2)A(1, 2), B(4,2)B(4, 2), and C(1,5)C(1, 5). Translate this triangle using the vector (−34)\begin{pmatrix} -3 \\ 4 \end{pmatrix} and find the coordinates of the image.

Solution:

  1. Identify the translation values: a=−3a = -3 (move 3 units left) and b=4b = 4 (move 4 units up).
  2. Apply the rule (x,y)→(x−3,y+4)(x, y) \rightarrow (x - 3, y + 4) to each vertex:
  • For A(1,2)A(1, 2): A′(1−3,2+4)=A′(−2,6)A'(1 - 3, 2 + 4) = A'(-2, 6)
  • For B(4,2)B(4, 2): B′(4−3,2+4)=B′(1,6)B'(4 - 3, 2 + 4) = B'(1, 6)
  • For C(1,5)C(1, 5): C′(1−3,5+4)=C′(−2,9)C'(1 - 3, 5 + 4) = C'(-2, 9)
  1. The coordinates of the image are A′(−2,6)A'(-2, 6), B′(1,6)B'(1, 6), and C′(−2,9)C'(-2, 9).

Explanation:

To translate a shape, we add the horizontal component of the vector to the xx-coordinates and the vertical component to the yy-coordinates of all vertices.

Problem 2:

Point PP is located at (3,−5)(3, -5). Find the coordinates of the image P′P' after a reflection in the yy-axis, followed by a rotation of 180∘180^\circ about the origin.

Solution:

Step 1: Reflect P(3,−5)P(3, -5) in the yy-axis. The rule for reflection in the yy-axis is (x,y)→(−x,y)(x, y) \rightarrow (-x, y). P(3,−5)→Ptemp(−3,−5)P(3, -5) \rightarrow P_{temp}(-3, -5).

Step 2: Rotate the new point Ptemp(−3,−5)P_{temp}(-3, -5) by 180∘180^\circ about the origin. The rule for a 180∘180^\circ rotation is (x,y)→(−x,−y)(x, y) \rightarrow (-x, -y). Ptemp(−3,−5)→P′(−(−3),−(−5))=P′(3,5)P_{temp}(-3, -5) \rightarrow P'(-(-3), -(-5)) = P'(3, 5).

Final Answer: P′(3,5)P'(3, 5).

Explanation:

This is a composite transformation. We apply the first rule (reflection) to the original point to get an intermediate point, then apply the second rule (rotation) to that intermediate point to find the final image.

Problem 3:

A square has vertices at J(2,2)J(2, 2), K(4,2)K(4, 2), L(4,4)L(4, 4), and M(2,4)M(2, 4). Reflect the square in the line y=xy = x. What are the coordinates of the vertices of the image J′K′L′M′J'K'L'M'?

A square on a coordinate plane with a diagonal reflection line y = x.

Solution:

Applying the rule for reflection in the line y=xy = x: (x,y)→(y,x)(x, y) \rightarrow (y, x). J(2,2)→J′(2,2)J(2, 2) \rightarrow J'(2, 2) K(4,2)→K′(2,4)K(4, 2) \rightarrow K'(2, 4) L(4,4)→L′(4,4)L(4, 4) \rightarrow L'(4, 4) M(2,4)→M′(4,2)M(2, 4) \rightarrow M'(4, 2)

Explanation:

Since the line y=xy = x passes through (2,2)(2,2) and (4,4)(4,4), these points remain fixed (invariant). The points (4,2)(4,2) and (2,4)(2,4) swap positions across the diagonal line.

Problem 4:

A point AA is at (−2,−1)(-2, -1). Rotate point AA by 90∘90^\circ clockwise about the origin (0,0)(0, 0) to find A′A', then translate A′A' using the vector (3−2)\begin{pmatrix} 3 \\ -2 \end{pmatrix} to find A′′A''. State the final coordinates of A′′A''.

The sequence of transformation from point A to A' and finally to A'' on a coordinate grid.

Solution:

Step 1: Rotate (−2,−1)(-2, -1) by 90∘90^\circ clockwise (same as 270∘270^\circ counter-clockwise). Rule: (x,y)→(y,−x)(x, y) \rightarrow (y, -x). A(−2,−1)→A′(−1,−(−2))=A′(−1,2)A(-2, -1) \rightarrow A'(-1, -(-2)) = A'(-1, 2).

Step 2: Translate A′(−1,2)A'(-1, 2) by (3−2)\begin{pmatrix} 3 \\ -2 \end{pmatrix}. (−1+3,2+(−2))=(2,0)(-1 + 3, 2 + (-2)) = (2, 0). Final coordinates A′′=(2,0)A'' = (2, 0).

Explanation:

First, the rotation moves the point from the third quadrant to the second quadrant. Then, the translation moves the point 3 units right and 2 units down to the x-axis.