Question

Given ΔBCD where B(2,1), C(3,4), and D(4,1), reflect over the line y=1.

Answer 1

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 $\underline{B'(2, 1), \ C'(3, -2), \ D'(4, 1)}$.

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 × $\mathbf{d_i}$

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 $\underline{B'(2, \, 1), \, C'(3, \, -2), \, D'(4, \, 1)}$

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Answer 2

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).

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