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A transformation is a general term for four specific ways to manipulate the shape of a point, a line or shape. The original shape of the object is called the final shape and the position of the object is the image under the transformation.
There is three fundamental transformation.
The reflection of a geometrical figure means the formation of the image of the figure after reflecting about the line of reflection. The line of reflection is also called the axis of reflection.
Properties of reflection
Very often reflections are performed using coordinates. The coordinates allow us to easily describe the image and its pre-image (object)
A reflection in the x-axis can be seen in the picture below. In which A is reflected its image 'A'. The general rule for the reflection in the x-axis:
A ( x, y) → A' (x, -y)
r_{x-axis}(x, y) = (x, -y)
A reflection in the y-axis can be seen in the picture below in which A is reflected its image A'. The general rule for a reflection in the line y=x:
A ( x, y) → A'(y, x) or
r_{y= x} (x, y) = (y, x)
When you reflect a point across the line y = -x, then x- coordinate and y-coordinate change places and are negated. The general rule for the reflection in the line y = -x:
A ( x, y) → A' (-Y, -X) or
r_{y = -x} (x, y) = (-y, -x)
A reflection in the line y = -x can be seen in the picture below in which A is reflected its image A'.
A perpendicular line is drawn from point A, B and C to the axis of reflection.The line from A be AP, from B be BR and from C be CQ. E longate lines AP,BR and CQ so that AP =PA', BR=RB' and CQ be QC'. Now join the points A', B' and C' and the formed \(\triangle\)A'BC' is the image of\(\triangle\) ABC in the axis m.
A perpendicular line is drawn from point A, B, C and D to the axis of reflection. The line from A be AP, from B be BQ and from C be CR and D be DS. E longate lines AP, BQ, CR and DS so that AP =PA', BQ=QB', CR=RC' and DS =SD'. Now join the points A', B',C' and D' and the formed \(\triangle\)A'BC' is the image of\(\triangle\) A'B'C'D' in the axis m.
A perpendicular line is drawn from point A, B, C,D and E to the axis of reflection. The line from A be AP, from B be BQ and from C be CR, D be DS and E be ET. E longate lines AP,BQ,CR,DS and ET so that AP =PA', BQ=QB', CR=RC', DS =SD' and ET=TE'. Now join the points A', B',C' D' and E' and the formed \(\triangle\)A'BC'D'E' is the image of\(\triangle\) in the axis m.
Solution:
Plot the points A (-2, 5); B (-2, -1); C (-5, -4); D (-5, 2) on the graph paper. Now join AB, BC, CD and DA to get a parallelogram.
When reflected in y-axis, we get A' (2, 5); B' (2, -1); C' (5, -4); D' (5, 2). Now join A'B', B'C', C'D' and D'A'.
Thus, we get the parallelogram A'B'C'D as the image of the parallelogram ABCD in y-axis.
Solution:
Image of the following points when reflected in y-axis.
(i) The image of (-3 , 3) is (3 , 3).
(ii) The image of (2, 4) is (-2, 4).
(iii) The image of (-2 , -6) is (2, -6).
(iv) The image of (5, -7) is (-5, -7).
Solution:
(i) The image of A(-7, 9) is A' (7, 9).
(ii) The image of B (-3, -6) is B' (3, -6).
(iii) The image of C(4, 8) is C' (-4, 8).
(iv) The image of D (5, -7) is D' (-5, -7).
Solution:
(i) The image of A(1, 4) is A’ (-1, -4).
(ii) The image of b (-3, -7) is B’ (3, 7).
(iii) The image of C (-5, 8) is C’ (5, -8).
(iv) The image of D (6, -2) is D’ (-6, 2).
Solution:
Plot the points X (1, -5); Y (6, -1); Z(-4, -3) on the graph paper. Now join XY, YZ and ZX; to get a triangle XYZ.
When reflected in x-axis, we get X' (1, 5); Y' (6, 1); Z' (-4, 3). Now join X'Y', Y'Z' and Z'X'.
Thus, we get a triangle X'Y'Z' as the image of the triangle XYZ in x-axis.
Solution:
(i)The image of (-5 , 2) is (-5 , -2).
(ii) The image of (2, -7) is (2, 7).
(iii) The image of (3, 4) is (3, -4).
(iv) The image of (-5, -4) is (-5, 4).
Solution:
The image of P (-6, -8) is P' (-6, 8).
The image of Q (4, 6) is Q' (4, -6) .
The image of R (-3, 2) is R' (-3, -2).
The image of S (3, -2) is S' (3, 2).
Solution:
When rotated through 90° about the origin in clockwise direction, the new position of the above points are;
(i) The new position of point P (3, 7) will become P' (7, -3)
(ii) The new position of point Q (-4, -7) will become Q' (-7, 4)
(iii) The new position of point R (-7, 5) will become R' (5, 7)
(iv) The new position of point S (2, -4) will become S' (-4, -2)
Solution:
P(x,y)=P'(x,-y)
A'(1,3)\(\rightarrow\)A'(1,-3)
B(4,5)\(\rightarrow\)B'(4,-5)
C(6,2)\(\rightarrow\)C'(6,-2)
Solution:
P(x,y)=P'(-x,y)
P(2,3)=P'(-2,3)
Q(4,5)=Q'(-4,5)
R(6,2)=R'(-6,2)
Solution:
P(x,y)\(\rightarrow\)P'(x,-y)
A(3,2)\(\rightarrow\)A'(3,-2)
B(5,6)\(\rightarrow\)B'(5,-6)
C(8,1)\(\rightarrow\)C'(8,-1)
Solution:
P(x,y)\(\rightarrow\)P'(-x,y)
A(-1,5)\(\rightarrow\)A'(1,5)
B(4,1)\(\rightarrow\)B'(-4,1)
C(-4,-3)\(\rightarrow\)C(4,-3)
A (1,2), B(4,5) and C(5,1) are vertices of \(\triangle\)ABC.Rotate \(\triangle\) ABC through 90^{o}.
Solution:
P(x,y)\(\rightarrow\)P'(y,-x)
A(1,2)\(\rightarrow\)A'(2,-1)
B(4,5)\(\rightarrow\)B'(5,-4)
C(5,1)\(\rightarrow\)C'(1,-5)
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