A reflection is a transformation that flips a figure across a line. The line work as a plane mirror. In reflection, the line joining the object and the image is perpendicular to the mirror line. It means the mirror line is perpendicular bisector of the line segment joining object and image. The mirror line is also called the axis of reflection.
When geometrical figures are reflected in the axis of reflection, the following properties are found.
a. Coordinates can be used for finding images of geometrical figures after the reflection in the lines like X- axis, Y- axis, a line parallel to X- axis, a line parallel to Y- axis, the line y = x, the line y = -x, etc.The distance of the object from the axis of reflection is equal to the distance of reflection is equal to the distance of the image from the axis is a reflection.
OP = OP' as shown in fig 1.
b. The shape of objects and images are laterally inverted. It means top remains at the top, bottom remains at the bottom but left side goes to the right side and right side goes to the left side as shown in fig 2.
c. The lines joining the same ends of the object and image are perpendicular to reflecting axis.
Axis of reflection is the perpendicular bisector of the line segment joining same ends of object and image.
XX'is perpendicular bisector of AA', BB' and CC' as in fig 3.
d. The points on the axis of reflection are invariant points.
e. Use of co-ordinates in Reflection.
f. In reflection, the object figure and its image figure are congruent to each other.
x = \(\frac {x+x'}2\) | and | 0 = \(\frac {y+y'}2\) |
or, x + x' = 2x | and | y + y' = 0 |
or, x' = x | and | y' = -y |
0 = \(\frac{x + x'}2\) | and | y = \(\frac {y + y'}2\) |
or, x + x' = 0 | and | y + y' = 2y |
or, x' = -x | and | y' = y |
x = \(\frac{x + x'}2\) | and | k = \(\frac{y + y'}2\) |
or, x + x' = 2x | and | y + y' = 2k |
or, x' = x | and | y' = 2k - y |
k = \(\frac {x + x'}2\) | and | y = \(\frac {y + y'}2\) |
or, x + x' = 2k | and | y + y' = 2y |
or, x' = 2k - x | and | y' = y |
Soln:
Here, the perpendiculars AP, BQ and CR are drawn from the points A, B and C on the line m respectively. Again, AP = PA^{1}, BQ = QB^{1}and CR = CR^{1} are drawn by producing Ab, BQ and CR respectively. Then A', B' and C' are joined . So, that the imageΔA'B'C' of triangle ABC is formed.
Soln:
As in the above figure, the perpendiculars AP, BQ, DR and CS are drawn in the mirror line 'm' from the points A, B, C and D respectively. Again AP = PA', BQ = QB', DR = RD' and CS = CS' are drawn by producing AP, BQ, QR and CS respectively. THen A',B',C' and D' are joined so that the image quadrilateral A'B'C'D' of the quadrilateral ABCD is formed.
Soln:
From the given points P and Q, perpendiculars PA and QB are drawn on the mirror line 'm'. Also, PA and QB are produced into P' and Q' so that PA = AP' and QB = BQ'. So the line P'Q' joining P' and Q' is the image of the line PQ.
Soln:
From the given figure, the mirror line 'm' is the perpendicular bisector of the line PQ. So the image P' and Q and image Q' of Q is at P. So, the image of PQ is Q'P' which is the same line PQ.
Soln:
Here, the mirror line 'm' is perpendicular to both the lines AD and BC. Let 'm' cuts AD at P and BC at Q. Now DP is produced up to D' so that DP = PD'. Similarly, APis produced up to A' so that AP = PA' BQ up to B' so that BQ = QB' and CQ up to C' so that CQ = QC'. Now the image A'B'C'D' of ABCD is formed.
Soln:
We have, reflection of a point on x-axis remains the x-coordinate same but the sign of y-coordinate changes. i.e. P(a, b)→ P'(a, -b).
So the images of the above points under the reflection on x - axis are:
A(3, 0)→A'(3, 0), B(4, -3)→ B'(4, 3), C(0, -7)→ C'(0, 7), D(-2, 6)→ D'(-2, -6), E(-3, -3)→ E'(-3, 3), F(-3, -9)→ F'(-3, 9), G(6, 6)→ G'(6, -6), H(7, -7)→ H'(7, 7). Ans
Soln:
We have the reflection of the point P(a, b) on the y-axis is P'(-a, b). i.e. P(a, b)→ P'(-a, b). So the images of the given points are:
A(3, 0)→ A'(-3, 0), B(4, -3)→ B'(-4, -3), C(0, -7)→ C'(0, -7), D(-2, 6)→ D'(2, 6), E(-3, -3)→ E'(3, -3), F(-3, -9)→ F'(3, -9), G(6, 6)→ G'(-6, 6) and H(7, -7)→ H'(-7, -7). Ans.
Soln:
We have, a reflection of a point P(a, b) on the line x = y is P'(b, a). i. e. P(a, b) → P'(b, a). So the images of the given points are:
A(3, 0)→ A'(0, 3), B(4, -3)→ B'(-3, 4), C(0, -7)→ C'(-7, 0), D(-2, 6)→ D'(6, -2), E(-3, -3)→ E'(-3, -3), F(-3, -9)→ F'(-9, -3), G(6, 6)→ G'(6, 6), H(7, -7)→ H'(-7, 7). Ans.
Soln:
We have, a reflection on x-axis is P(a, b)→ P'(a, -b). So reflection of triangle PQR on x-axis given the images coordinates as:
P(1, 1) →P'(1, -1)
Q(3, 1)→ Q'(3, -1)
R(3, -1)→ R'(3, 1)
Here, the shaded portion triangle P'Q'R' is the image of triangle PQR.
Soln:
We have, reflection on the line y = -x is P(a, b)→ P'(-b, -a). So, we have,
P(1, 1) →P(-1, -1)
Q(3, 1)→ Q'(-1, -3)
R(3, -1)→ R'(1, -3)
The graph is given below:
The shaded portion triangle P'Q'R' is the image of triangle PQR.
Soln:
We have, a reflection on the line y = k is P(a, b)→ P'(a, 2k - b) where k = 3. So,
P(1, 1)→ P'(1, 2×3 - 1) = P'(1, 5)
Q(3, 1)→ Q'(3, 2× 3 -1) = Q'(3, 5)
R(3, -1)→ R[3, 2× 3 - (-1)] = R'(3, 7)
The graph is as shown below:
The shaded portion triangle P'Q'R' is teh image of triangle PQR.
Soln:
Here, the sign of y-coordinate of the point A(3, 4) is changed in the image A'(3, -4) under the reflection R. So the axis of reflection is x-axis.
Soln:
Here, Y reflects the point of A(3, 4) to A'(-1, 4), is which y-coordinate is same but x-coordinate changes from 3 to -1. So the line AA' is parallel to x-axis. Here, the mirror line is the perpendicular bisector of AA' or the mirror line is parallel to y-axis which passes through the mid-point of AA' i. e. \(\frac{3-1}{2}\), \(\frac{4+4}{2}\) = (1, 4). So, the line parallel to y-axis (1, 4) is x = 1.
∴ Axis of reflection is x = 1. Ans
Soln:
Here, reflection R reflects the point A(4, 5) to A'(-5, -4). So, x and y coordinates are interchanged and the signs are also changed. i. e. P(a, b)→ P'(-b, -a). So the axis of reflection is the line y = -x. Ans.
Soln:
Here, reflection P reflects the point A (-2, 3) to A'(6, 3) is which y-coordinate is same but x-coordinate changes from -2 to 6. In this case as in the reflection of the line x = h, P(a, b)→ P'(2h-a, b)
A(-2, 3)→ A'(6, 3). So,
2h - a = 6
or, 2h -(-2) = 6
or, 2h + y = 6
or, 2h + y = 6
or, 2h = 6 - 2 = 4
∴ h = 2
∴ The axis of reflection is h = 2 which is parallel to y-axis.
Soln:
Here, reflection Q reflects the point A(3, 4) to A'(3, 2) is which x-coordinate is same but y-coordinate changes from 4 to 2. In this axis the reflection of the line y = k, P(a, b)→ P'(a, 2k - b),
A(3, 4)→ A'(3, 2).
so, 2k - b = 2
or, 2k - 4 = 2
or, 2k = 2 + 4 = 6
∴ k = 3
∴ The axis of reflection is the line k = 3 which is parallel to x-axis. Ans.
Soln:
i)Reflection of the given trapezium PQRS under, x-axis is as
P(a, b) → P'(a, -b) we get,
P(2, 1)→ P'(2, -1)
Q(1, -2)→ Q'(1, 2)
R(-3, -2)→ R'(-3, 2) and
S(-5, 1)→ S'(-5, -1)
The graph of trapezium PQRS and itsn image P'Q'R'S' is as shown below:
ii)Reflection of trapezium on y-axis as P(a, b)→ P'(-a, b) we get,
P(2, 1)→ P'(-2, 1)
Q(1, -2)→ Q'(-1, -2)
R(-3, -2)→ R'(3, -2) and S(-5, 1)→ S'(5, 1)
The graph of trapezium PQRS and its image P'Q'R'S' is shown below:
Find the image of the point (-3,-4) after reflection in the line.
x=4
(1,4)
(8,-3)
(7,-1)
(11,-4)
Find the image of the point (-3,-4) after reflection in the line.
x=-3
(3,4)
(-9,2)
(-3,-4)
(-2,4)
Find the image of the point (-3,-4) after reflection in the line.
y=3
(-3,5)
(-3,10)
(-1,1)
(1,2)
Find the image of the point (-3,-4) after reflection in the line.
y=-2
(-6,8)
(-9,8)
(-3,0)
(-2,1)
The image of A (a,5-b) when reflected in X-axis is A'(2a-3,2b-8).Find the value of a and b.
3,3
2,2
4,4
1,1
The image of A (a-2,3b) when reflected in X-axis is A'(2a-5,b-12).Find the value of a and b.
2,2
3,3
1,1
4,4
The image of P(2,6-b) when reflected in Y-axis is P'(4a-6,b+2).Find the value of a and b.
2,3
1,2
3,4
4,5
The image of P(3-a,4-b) when reflected in Y-axis is P'(2a-5,b+2).Find the value of a and b.
2,1
3,3
2,2
1,1
The image of Q(3a, 4b) when reflected in the line x=y is Q'(5b-2, 4a-3).Find the co-ordinates of Q.
The image of P(2a, 3b) when reflected in the line x=y is P'(4b-1, 3a-2).Find the co-ordinates of P.
P(6,1)
P(1,2)
P(4,3)
P(7,8)
The image of R(3-a,b) when reflected in the line x+y=0 is R'(2b-6, 2a-5).Find the co-ordinates of R.
R(1,2)
R(7,7)
R(5,5)
R(8,3)
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