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Geometry and Measurement - Transformational Geometry (Translation, Reflection, Rotation)

Grade 8IB

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

🔑Concepts

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A translation moves every point of a figure the same distance in the same direction. It is often described by a vector \begin{pmatrix} a \\ b \\end{pmatrix}, where aa is the horizontal shift and bb is the vertical shift. The mapping is (x,y)→(x+a,y+b)(x, y) \to (x + a, y + b). Orientation and size remain unchanged.

A triangle translated 2 units right and 1 unit up.
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A reflection creates a mirror image of a figure across a line of reflection. Points on the line remain fixed (invariant), while all other points are 'flipped' such that the line is the perpendicular bisector of the segment connecting a point and its image.

A triangle reflected across the x-axis.
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A rotation turns a figure about a fixed point called the center of rotation. A positive angle indicates counter-clockwise (CCW) rotation, while a negative angle indicates clockwise (CW) rotation. The distance from the center to any point remains constant.

A triangle rotated 90 degrees counter-clockwise about the origin.
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Invariant points are points that do not move during a transformation. For example, during a reflection, any point on the line of reflection is invariant. During a rotation, the center of rotation is invariant.

📐Formulae

Translation rule: (x,y)→(x+a,y+b)(x, y) \to (x + a, y + b)

Reflection over the x-axis: (x,y)→(x,−y)(x, y) \to (x, -y)

Reflection over the y-axis: (x,y)→(−x,y)(x, y) \to (-x, y)

Reflection over the line y=xy = x: (x,y)→(y,x)(x, y) \to (y, x)

Reflection over the line y=−xy = -x: (x,y)→(−y,−x)(x, y) \to (-y, -x)

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

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

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

💡Examples

Problem 1:

Triangle ABCABC has vertices A(1,2)A(1, 2), B(4,2)B(4, 2), and C(1,5)C(1, 5). Apply a translation using the vector (−31)\begin{pmatrix} -3 \\ 1 \\ \end{pmatrix} and then reflect the resulting image over the x-axis. Find the final coordinates of vertex C′′C''.

Solution:

Step 1: Apply the translation to point C(1,5)C(1, 5). Using the rule (x+a,y+b)(x + a, y + b), we get C′(1+(−3),5+1)=C′(−2,6)C'(1 + (-3), 5 + 1) = C'(-2, 6). Step 2: Apply the reflection over the x-axis to C′(−2,6)C'(-2, 6). Using the rule (x,−y)(x, -y), we get C′′(−2,−6)C''(-2, -6). The final coordinates of C′′C'' are (−2,−6)(-2, -6).

Explanation:

To find the final position, we apply the transformations sequentially. The translation shifts the point 3 units left and 1 unit up. The reflection over the x-axis then negates the y-coordinate while keeping the x-coordinate the same.

Problem 2:

A square has a vertex at S(2,−3)S(2, -3). If the square is rotated 90∘90^\circ counter-clockwise about the origin, what are the coordinates of the image S′S'?

Solution:

Step 1: Identify the starting coordinates (x,y)=(2,−3)(x, y) = (2, -3). Step 2: Apply the rotation rule for 90∘90^\circ counter-clockwise, which is (x,y)→(−y,x)(x, y) \to (-y, x). Step 3: Substitute the values: x′=−(−3)=3x' = -(-3) = 3 and y′=2y' = 2. The coordinate of S′S' is (3,2)(3, 2).

Explanation:

A 90∘90^\circ counter-clockwise rotation swaps the x and y values and changes the sign of the original y-coordinate. Visually, the point moves from Quadrant IV to Quadrant I.

Problem 3:

Rectangle PQRSPQRS has vertices P(1,1)P(1, 1), Q(3,1)Q(3, 1), R(3,2)R(3, 2), and S(1,2)S(1, 2). Reflect the rectangle across the line y=xy = x. Determine the coordinates of the image P′Q′R′S′P'Q'R'S'.

A rectangle reflected across the line y=x.

Solution:

The rule for reflection across the line y=xy = x is (x,y)→(y,x)(x, y) \to (y, x). Applying this to each vertex: P(1,1)→P′(1,1)P(1, 1) \to P'(1, 1) Q(3,1)→Q′(1,3)Q(3, 1) \to Q'(1, 3) R(3,2)→R′(2,3)R(3, 2) \to R'(2, 3) S(1,2)→S′(2,1)S(1, 2) \to S'(2, 1)

The new coordinates are P′(1,1)P'(1, 1), Q′(1,3)Q'(1, 3), R′(2,3)R'(2, 3), and S′(2,1)S'(2, 1).

Explanation:

In a reflection over y=xy = x, the x-coordinates and y-coordinates of every point are swapped. Notice that point PP lies on the line y=xy=x, so it is an invariant point and does not change its position.

Problem 4:

Point M(2,4)M(2, 4) is rotated 180∘180^\circ about the origin to M′M', and then M′M' is translated by the vector \begin{pmatrix} 3 \\ -2 \\end{pmatrix} to reach point M′′M''. Find the coordinates of M′′M''.

Graph showing point M, its 180 degree rotation M', and the subsequent translation to M''.

Solution:

Step 1: Rotate M(2,4)M(2, 4) 180∘180^\circ about the origin. The rule is (x,y)→(−x,−y)(x, y) \to (-x, -y). M(2,4)→M′(−2,−4)M(2, 4) \to M'(-2, -4)

Step 2: Translate M′(−2,−4)M'(-2, -4) by \begin{pmatrix} 3 \\ -2 \\end{pmatrix}. The rule is (x,y)→(x+3,y−2)(x, y) \to (x + 3, y - 2). M′(−2,−4)→M′′(−2+3,−4−2)=M′′(1,−6)M'(-2, -4) \to M''(-2 + 3, -4 - 2) = M''(1, -6)

The final coordinates are M′′(1,−6)M''(1, -6).

Explanation:

A 180∘180^\circ rotation negates both coordinates. The subsequent translation adds the vector components to the results of the rotation.