A reflection about a line gives an image that is at an equal distance from the line but on the opposite (other) side of the line
The distance between the lines m and n is 6 units
Reason:
The given parameters are;
[tex]T_{<0, -12>}\Delta(XYZ) = (R_n \circ R_m) \Delta (XYZ)[/tex]
Let Y represent the top vertex of Δ(XYZ), and the coordinates of point Y is (3, 0) we have;
The coordinates of Y following [tex]T_{<0, -12>}[/tex] is Y''(3, -12)
A reflection of Y across the line m, y = -9, gives the coordinates of the image at Y'(3, -18)
A reflection of the image Y'(3, -18) across the line n, y = -15, gives the image at Y''(3, -12)
The distance between lines m, and line n, is -9 - (-15) = 6
Generally, the difference between two lines of reflection, n, and m, following a composite reflection is 2×(n - m)
Given that the total transformation is -12, we have;
2×(n - m) = -12
Therefore;
[tex](n - m) = \dfrac{-12}{2} = -6[/tex]
Which gives;
m - n = 6
The distance between the two lines = 6 units
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