Determine the coordinates of Q(6,-4) after a reflection in the line x=2

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JeanaShupp JeanaShupp · Answer 1 · 2020-11-08T03:02:15+01:00

Answer: Q'(-4,-4)

Step-by-step explanation:

Every line of equation x=a is parallel to y-axis.

Reflection across y axis rule :

(x,y) → (-x, y) i.e. y -coordinate remains same.

Reflection across x=a rule :

(x,y) → (a-x, y)

To determine the coordinates of Q(6,-4) after a reflection in the line x=2.

So, Q(6,-4) → Q' (2-6, -4)= Q'(-4,-4)

Hence, the required coordinate =Q'(-4,-4)

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MrRoyal MrRoyal · Answer 2 · 2021-11-11T11:12:03+01:00

When a point is reflected, it must be reflected across a line

The coordinate of Q after the reflection is: [tex]\mathbf{Q' = (- 2, -4)}[/tex]

So, Q(6,-4) → Q' (2-6, -4)= Q'(-4,-4)

The point is given as:

[tex]\mathbf{Q = (6,-4)}[/tex]

The line of reflection is given as:

[tex]\mathbf{x = 2}[/tex]

The rule of reflection for [tex]\mathbf{x = a}[/tex] is:

[tex]\mathbf{(x,y) \to (2a - x, y)}[/tex]

For [tex]\mathbf{a = 2}[/tex], we have:

[tex]\mathbf{(x,y) \to (2(2) - x, y)}[/tex]

[tex]\mathbf{(x,y) \to (4 - x, y)}[/tex]

So, we have:

[tex]\mathbf{Q' = (4 - 6, -4)}[/tex]

[tex]\mathbf{Q' = (- 2, -4)}[/tex]

Hence, the coordinate of Q after the reflection is: [tex]\mathbf{Q' = (- 2, -4)}[/tex]