Urgent need the answers plz help.
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Answer:
(a) [tex]P" = (-4,-3)[/tex]
(b) [tex](x,y) \to (4,-8)[/tex]
Step-by-step explanation:
Given
[tex]P = (4,3)[/tex]
Solving (a): Reflect across x and y-axis.
Reflection across x-axis has the following rules
[tex](x,y) \to (x,-y)[/tex]
So, we have:
[tex]P' = (4,-3)[/tex]
Reflection across y-axis has the following rules
[tex](x,y) \to (-x,y)[/tex]
So, we have:
[tex]P" = (-4,-3)[/tex]
Hence, the new point is: (-4,-3)
Solving (b): Rx . Do,2 (2,4)
[tex]R_x \to[/tex] reflect across the x-axis
Reflection across x-axis has the following rules
[tex](x,y) \to (x,-y)[/tex]
So, we have:
[tex](2,4) = (2,-4)[/tex] ---- when P is reflected across the x-axis
[tex]D_{o,2} \to[/tex] dilate by a scale factor of 2
The rule is:
[tex](x,y) \to 2 * (x,y)[/tex]
So, we have
[tex](x,y) \to 2 * (2,-4)[/tex]
Open bracket
[tex](x,y) \to (4,-8)[/tex]