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Given: [tex]\(P' = (1, -6)\)[/tex]

Transformation: [tex]\((x, y) \rightarrow (x-3, y-3)\)[/tex]

Reflection line: [tex]\(y = x\)[/tex]

Find the coordinates of [tex]\(P\)[/tex].

The coordinates of [tex]\(P\)[/tex] are [tex]\(\square\)[/tex].


Sagot :

To find the original coordinates [tex]\( P \)[/tex] given the transformation and reflection information, follow these steps:

1. Starting Point:
We start with the point [tex]\( P' \)[/tex] given as [tex]\( P' = (1, -6) \)[/tex].

2. Reflection Across the Line [tex]\( y = x \)[/tex]:
When a point [tex]\( (a, b) \)[/tex] is reflected across the line [tex]\( y = x \)[/tex], the coordinates are swapped. Hence, reflecting [tex]\( P' \)[/tex]:
[tex]\[ P'' = (-6, 1) \][/tex]

3. Reverse the Translation:
The transformation mentioned is [tex]\( (x, y) \rightarrow (x-3, y-3) \)[/tex], which means moving each x-coordinate and y-coordinate 3 units to the left and down, respectively. To find the original point before this transformation, we need to reverse it:
[tex]\[ (x', y') \rightarrow (x' + 3, y' + 3) \][/tex]
Applying this to the reflected point:
[tex]\[ P = (-6 + 3, 1 + 3) = (-3, 4) \][/tex]

Thus, the coordinates of [tex]\( P \)[/tex] are [tex]\( \boxed{-3, 4} \)[/tex].

Answer:

P = (-3, 4)

Step-by-step explanation:

Given:

  • [tex]\(P' = (1, -6)\)[/tex]
  • Reflection line: [tex]\(y = x\)[/tex]
  • Transformation: [tex]\((x, y) \rightarrow (x-3, y-3)\)[/tex]

Where, y = x

P' = (x, y) = (1, -6)

P'' = (-6, 1)

Using reverse transformation.

(x', y') = (x' + 3, y' + 3)

(x', y') = (-6 + 3, 1 + 3)

P = (-3, 4)

Therefore, the coordinates of P are (-3, 4) in the (x, y) format