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Mathematics 6.

Table of contents
Divisibility
How to solve problems?
Rational numbers I.
Axial symmetry
Axial symmetry in your environment
Lines of symmetry in triangles
Axially symmetric polygons and circles
Using the ruler and compass
Constructing perpendicular lines
Constructing parallel lines
Constructing, copying and bisecting angles
Constructing the images of figures under an axial symmetry
Constructing axially symmetric polygons
Mixed exercises
Rational numbers II.
Proportion
Percentages
Probability, statistics
Supplementary material
Constructing the images of figures under an axial symmetry
Example 1
There is a line of symmetry \latex{ l } and a point \latex{ P }. Construct the mirror image of point \latex{ P (P') } across line \latex{ l }.
Solution
Use the fact that the mirror image of point \latex{ P } is found on a line drawn from \latex{ P } perpendicular to line \latex{ l }, and points \latex{ P } and \latex{ P' } are found at the same distance from line \latex{ l }.
Steps of construction:
When reflecting across an axis, the image of point \latex{P} is a point \latex{P’} that is \latex{PP’\perp l, PL = P’L.}
  1. Construct line \latex{ l } and point \latex{ P }.
  2. From point \latex{ P }, construct a line perpendicular to line \latex{ l }.
  3. From point \latex{ L }, measure the distance \latex{ PL } on the perpendicular line to get point \latex{ P' }.
\latex{P}
\latex{L}
\latex{Q}
\latex{P’}
\latex{R}
\latex{l}
The perpendicular
bisector of segment
\latex{PP’} is line \latex{l}.
If point \latex{ P } is on line \latex{ l }, then its mirror image is itself, that is, \latex{ P' = P }.
 
Note:
If the compass is opened to the same radius during construction, the intersection of the arcs will be at point \latex{ P' } (you do not have to measure the \latex{ PL } distance). Since the \latex{ PSP'R } quadrilateral is a rhombus, its diagonals are the perpendicular bisectors of each other.
\latex{l}
\latex{R}
\latex{P}
\latex{P’}
\latex{L}
\latex{S}
Example 2
There is a line of symmetry \latex{ l } and a line segment \latex{ AB }. Construct the mirror image of line segment \latex{ AB } (line segment \latex{ A'B' }) across the axis \latex{ l }.
Solution
Use the fact that a line segment is clearly defined by its two endpoints.
Therefore, reflect the two endpoints of line segment \latex{ AB } and connect their reflections.
Steps of construction:
When reflected across an axis, the mirror image of a line segment is another line segment of equal length.
  1. Draw line \latex{ l } and line segment \latex{ AB }.
  2. Construct the mirror image of point \latex{ A } (point \latex{ A' }) across line \latex{ l }.
  3. Construct the mirror image of point \latex{ B } (point \latex{ B' }) across line \latex{ l }.
  4. Connect points \latex{ A' } and \latex{ B' }.
\latex{A}
\latex{B}
\latex{B’}
\latex{A’}
\latex{l}
Is it true?
A line segment and its mirror image are always found in different half-planes.
\latex{l}
\latex{\alpha}
\latex{\alpha}
Note:
The lines passing through line segments \latex{ AB } and \latex{ A'B' } intersect line \latex{ l } at the same point and form identical angles with it.
Example 3
There is a line \latex{ l } and an \latex{ ABC } triangle. Construct the mirror image of triangle \latex{ ABC } (triangle \latex{ A'B'C' }) across axis \latex{ l }.
Solution
Use the fact that a triangle is clearly defined by its three vertices. Therefore, reflect the vertices and connect them.
Steps of construction:
  1. Construct line \latex{ l } and triangle \latex{ ABC }.
  2. Construct the mirror images of the vertices \latex{(A’, B’, C’)}.
  3. Connect points \latex{A’, B’} and \latex{C’}.
\latex{C}
\latex{B}
\latex{B’}
\latex{C’}
\latex{l}
\latex{A}
\latex{A’}


Triangle \latex{ A'B'C' } is the mirror image of triangle \latex{ ABC; } thus, their corresponding angles are equal.

\latex{l}
\latex{\alpha}
\latex{\alpha}
\latex{\beta}
\latex{\beta}
\latex{\gamma}
\latex{\gamma}
\latex{C}
\latex{C’}
\latex{B’}
\latex{A’}
\latex{A}
\latex{B}
Orientation:
Note:
Reflection across a line of symmetry changes the orientation of the triangle. For example, if the vertices of triangle ABC follow each other in an anti-clockwise direction, then the vertices of the mirror image will follow each other in a clockwise direction.
\latex{C}
\latex{B}
\latex{A}
\latex{A’}
\latex{B’}
\latex{C’}
\latex{-}
\latex{+}
\latex{l}
Example 4
There is a line \latex{ l } and a circle with a centre \latex{ O } and a radius \latex{ r }. Construct the mirror image of the circle across line \latex{ l }.
Solution
The mirror image of a circle is a congruent circle with the same radius \latex{ (r' = r) }. Thus, it is enough to reflect the centre of the circle.
Steps of construction:
  1. Construct a line \latex{ l } and a circle with centre \latex{ O } and radius \latex{ r }.
  2. Construct the mirror image of the centre.
  3. Draw a circle with a radius \latex{ r } around centre \latex{ O'. }
\latex{c}
\latex{O}
\latex{c’}
\latex{O’}
\latex{l}
\latex{r}
\latex{r’}
\latex{l}
\latex{A}
\latex{B=B’}
\latex{A’}
\latex{O}
\latex{c=c’}
\latex{C=C’}
Note:
Since the circle is axially symmetric, if one of its diagonals is on line \latex{ l }, then the mirror image of the circle is itself, even though only two points of the circle remain in the same place.
Exercises
{{exercise_number}}. Construct the mirror images of the line segments shown below across line \latex{ l }. What can you say about the line segments and their reflections? \latex{(EF \parallel l; l} is the perpendicular bisector of line segment \latex{GH; IJ \perp l.)}
\latex{l}
\latex{A}
\latex{B}
\latex{C}
\latex{D}
\latex{E}
\latex{F}
\latex{G}
\latex{H}
\latex{I}
\latex{J}
{{exercise_number}}. Construct the mirror images of the lines shown below across line \latex{ l }. What can you say about the lines and their reflections? \latex{(g \parallel l.)}
\latex{l}
\latex{e}
f
\latex{g}
{{exercise_number}}. Construct the mirror images of the rays shown below across line \latex{ l }. What can you say about the rays and their reflections?
\latex{l}
\latex{a}
\latex{b}
\latex{c}
\latex{A}
\latex{B}
\latex{C}
{{exercise_number}}. Construct the mirror images of the circles shown below across line \latex{ l }. What can you say about the circles and their reflections?
\latex{l}
\latex{A}
\latex{B}
\latex{E}
\latex{F}
\latex{C}
\latex{G}
\latex{H}
\latex{D}
\latex{I}
{{exercise_number}}. Construct the mirror images of the triangles shown below across line \latex{ l }. What can you say about the triangles and their reflections?
\latex{l}
\latex{A}
\latex{B}
\latex{C}
\latex{D}
\latex{F}
\latex{E}
\latex{H}
\latex{G}
\latex{I}
\latex{J}
\latex{L}
\latex{K}
{{exercise_number}}. Construct angle \latex{\alpha = 60°}. Draw a line \latex{ l }, which intersects both arms of angle \latex{\alpha}. Construct the reflection of angle \latex{\alpha} across line \latex{l}.
{{exercise_number}}. Endpoint \latex{ A } of line segment \latex{ AB } is on the line of symmetry. Construct the mirror image of the line segment. What polygon is formed by points \latex{ ABB'A' }?
{{exercise_number}}. Line segment \latex{ AB } has no common points with the line of symmetry. Construct the mirror image of the line segment. What polygon is formed by points \latex{ ABB'A' }?
{{exercise_number}}. Construct a triangle that has a right angle and legs that are \latex{ 3 } and \latex{ 5\, cm } long. Reflect the triangle
  1. across the hypotenuse;
  1. across one of the legs.
What type of polygon is formed by the original triangle and its mirror image?
{{exercise_number}}. Determine the coordinates of the vertices of triangle \latex{ABC} shown in the coordinate system below. 
  1. Reflect triangle \latex{ ABC } across the \latex{ y }-axis. Determine the coordinates of the resulting \latex{ A'B'C' } triangle's vertices.
  2. Reflect triangle \latex{ ABC } across the \latex{ x }-axis as well. Determine the coordinates of the resulting \latex{ A'B'C' } triangle's vertices.
\latex{C}
\latex{B}
\latex{A}
\latex{-15}
\latex{-14}
\latex{-13}
\latex{-12}
\latex{-11}
\latex{-10}
\latex{-9}
\latex{-8}
\latex{-7}
\latex{-6}
\latex{-5}
\latex{-4}
\latex{-3}
\latex{-2}
\latex{-1}
\latex{-1}
\latex{-2}
\latex{-3}
\latex{-4}
\latex{-5}
\latex{-6}
\latex{6}
\latex{5}
\latex{4}
\latex{3}
\latex{2}
\latex{1}
\latex{1}
\latex{2}
\latex{3}
\latex{4}
\latex{5}
\latex{6}
\latex{7}
\latex{8}
\latex{9}
\latex{10}
\latex{11}
\latex{12}
\latex{13}
\latex{14}
\latex{15}
\latex{x}
\latex{y}
{{exercise_number}}. There is a point \latex{ P } and a point \latex{ P' }. Point \latex{ P' } is the mirror image of point \latex{ P }. Construct the line of symmetry.
Quiz
 
Plot point \latex{ A }\latex{ (5; 2) } in a Cartesian coordinate system. Reflect point A across the \latex{ x }-axis, then reflect the mirror image across the \latex{ y }-axis. After that, reflect the resulting mirror image across the \latex{ x }-axis, and so on.
What are the coordinates of the points after the \latex{a) 2,006th;\quad b) 2,007th;\quad c) 2,008th;\quad d) 2,009th} reflection?

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