cadaVR anatomy by Mozaik Education

Rotation about a point in the plane

Let us take two straight lines with a \latex{ 60º } angle included between them (let them be denoted by l₁ and l₂, let their intersection point be O) and triangle ABC. Let us reflect this triangle about straight line l₁, then let us reflect the image (triangle A’B’C’) about straight line l₂. (Figure 45)

Figure 45

\latex{ A}

\latex{ C}

\latex{ B}

\latex{ B' }

\latex{ C' }

\latex{ A' }

\latex{ A }

\latex{ B}

\latex{ C}

\latex{l_{1} }

\latex{l_{2} }

\latex{O }

\latex{60°}

When considering the relative positions of triangle A’’B’’C’’ resulting after the second reflection and of the original triangle ABC and using the properties of the line reflection we can realise the following:

\latex{OA=OA}
\latex{AOA}

Based on the above example doing the two line reflections in succession defines a new congruent transformation in the plane: rotation about a point.

DEFINITION: Point O of the plane and directed angle \latex{\alpha} are given, regarding the measure of the latter one: \latex{0°\leq \left| \alpha\right| \leq 360°} (Figure 46). Let us assign point P' to every single point P of the plane as follows:

we assign O to itself, i.e. O = O’;
if \latex{P\neq O}, then P’ is the point of the plane for which OP = OP’, and ray OP’ is the rotated image of ray OP when rotated through directed rotation angle \latex{\alpha}.

Point O is the centre of rotation.

Rotation about a point is unambiguously defined by point O and a directed angle \latex{\alpha}, or point O and point P different from point O with its image point P'.

Example 1

Let us take triangle ABC and point O. Let us rotate the triangle about O through \latex{ 90º }.

Solution

The rotated images of the vertices of the triangle unambiguously define the image triangle. (Figure 47)

Figure 47

\latex{ A }

\latex{ B }

\latex{ C }

\latex{ B' }

\latex{ C' }

\latex{ A' }

\latex{ O }

The properties of rotation about a point

If a \latex{\alpha \neq 0°\left(\alpha \neq 360°\right) }, then the sole fixed point of the transformation is the centre O of the rotation. If \latex{\alpha =0°\left(\alpha =360°\right) }, then every point of the plane is a fixed point (identical transformation). (Figure 48)
If \latex{\alpha} = ±180º, then the rotation about \latex{ O } is the same as point reflection in O. (Figure 49)
Rotation about a point is a distance-preserving and angle-preserving transformation. (Figure 50)
Rotation about point \latex{ O } through directed angle \latex{\alpha}can be produced by doing two line reflections in succession where the intersection point of the axes is point O and the angle included between the axes is \latex{\frac{\left|\alpha \right| }{2} }. (Figure 51)

Figure 51

\latex{ t_{1} }

\latex{ t_{2} }

\latex{ t_{1} }

\latex{\alpha}

\latex{-\alpha}

\latex{P}

\latex{P'}

\latex{P}

\latex{P'}

The direction of rotation is defined by the order of the line reflections. (It can be proven that in the case of a given directed angle a this production is independent of the direction of the axes, i.e. one of the axes can freely be chosen, but the other axis is unambiguously defined by the choice of the first axis.)

The previous property (using that line reflection is an orientationchanging geometric transformation) implies that rotation about a point is an orientation-preserving geometric transformation.

Exercises

{{exercise_number}}. Take a triangle with \latex{ 3, 4 } and \latex{ 5 } \latex{ cm } long sides. Rotate this triangle about the vertex opposite the shortest side

through \latex{ +45º };

through \latex{ +90º };

through \latex{ –60º };

through \latex{ –180º };

through \latex{ +270º };

through \latex{ –90º }.

{{exercise_number}}. Take a \latex{ 60º } angle and point O in the angular domain. Rotate the angle about O

through \latex{ 30º };

through \latex{ –60º };

through \latex{ –45º }.

{{exercise_number}}. Take points A and B. Construct the set of points about which you can rotate point A so that the image will be point B.

{{exercise_number}}. Take two non-parallel line segments of equal length. Construct a point about which you can rotate one of the line segments so that the image will be the other line segment.

{{exercise_number}}. Take points A and B. Construct the centre of rotation which transforms point A to point B if the angle of rotation is:

\latex{ +90º };

\latex{ –90º };

\latex{ +60º };

\latex{ +120º }.

{{exercise_number}}. In the Cartesian coordinate system the vertices of a triangle are: A(–1; 1), B(4; 3), C(–3; 5). Rotate the triangle about the origin

through \latex{ +90º };

through \latex{ –90º };

through \latex{ +180º };

through \latex{ +270º }.

Give the coordinates of the vertices of the image triangle in all the cases.

{{exercise_number}}. In the Cartesian coordinate system the vertex at the right angle in an isosceles right-angled triangle is the origin, and one end-point of its hypotenuse is point

(\latex{ 1; 1 });

(\latex{ 3; 4 });

(\latex{ 4; –1 });

(\latex{ –3; –8 }).

Determine the coordinates of the other end-point of the hypotenuse in each case.

{{exercise_number}}. Take two parallel straight lines and a point between them. Construct a regular triangle so that one of its vertices is the given point and the other two vertices lie on the given straight lines each.

{{exercise_number}}. Take two points. Construct a square so that one of the given points is one of its vertices and the other given point is the intersection point of its diagonals.

{{exercise_number}}. The right-angled trapezium drawn in the square shown in the figure is rotated four times in succession about the centre of the square through \latex{ +90º }. Which parts of the square will be overlaid at least twice?

Mathematics 9.

-