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Sagot :
Points on ΔBCD located on the reflection line will form an image at the
same location.
Response:
- The image of ΔBCD is ΔB'C'D' where [tex]\underline{B'(2, 1), \ C'(3, -2), \ D'(4, 1)}[/tex].
Method for finding the image of a reflected object
The given points on ΔBCD are; B(2, 1), C(3, 4), and D(4, 1).
The line over which the points are reflected is; y = 1
Required:
The image of ΔBCD
Solution:
The line of reflection, y = 1, is parallel to the x-axis, therefore;
- The x-coordinate of the preimage = The x-coordinate of the image
- The vertical distance of each point on the preimage from the line of reflection, d = The vertical distance of points on the image from the line of reflection
Therefore;
y-coordinate of image = y-coordinate of preimage - 2 × [tex]\mathbf{d_i}[/tex]
The y=coordinate at point B = 1
Therefore, at point B, we have; d = 1 - 1 = 0
y-coordinate of point B' = 1 - 2 × 0 = 1
The coordinates of point B' = B'(2, 1)
At point C, d = 4 - 1 = 3
y-coordinate of point C' = 4 - 2×3 = -2
The coordinates of point C' = C'(3, -2)
The vertical distance of the point D from the line of reflection, d = 1 - 1 = 0
The y-coordinate of the point D' = 1 - 2 × 0 = 1
The coordinates of the point D' = (4, 1)
Therefore;
- The image of ΔBCD is ΔB'C'D' with vertices [tex]\underline{B'(2, \, 1), \, C'(3, \, -2), \, D'(4, \, 1)}[/tex]
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The image of ΔBCD on the line y=1 is ΔB'C'D' with vertices at points; B'(2, 1), C'(3, -2) and D' = (4, 1).
Determining the image of a reflected object
The vertices of triange ΔBCD are; B(2, 1), C(3, 4), and D(4, 1).
According to the question;
- The points are reflected is; y = 1
The image of ΔBCD will therefore be triangle ΔB'C'D' and it's vertices can be evaluated as follows;
The line of reflection, y = 1, is parallel to the x-axis, therefore;
In essence, The x-coordinate of the triangle = The x-coordinate of the image
The vertical distance of each point on the preimage from the line of reflection, d = The vertical distance of points on the image from the line of reflection
Therefore;
- The y=coordinate at point B = 1
- Since, the line is reflected on point y=1, the y-ordinate of point B remains constant.
The coordinates of point B' = B'(2, 1)
The y-ordinate of image = -2 × distance, i of image to the reflection line
At point C, i = 4 - 1 = 3
- For the y-coordinate of point C' = 4 +(-2×3) = -2
The coordinates of point C' = C'(3, -2)
The vertical distance of the point D from the line of reflection, i = 1 - 1 = 0
The y-coordinate of the point D' = 1 - 2 × 0 = 1
The coordinates of the point D' = (4, 1)
Therefore;
The image of ΔBCD on the line y=1 is ΔB'C'D' with vertices at points; B'(2, 1), C'(3, -2) and D' = (4, 1).
Read more on reflection over a line;
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