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Y_Kistochka [10]
3 years ago
5

rotation of 90 degrees counterclockwise about the origin, point O, then a reflection across the x-axis reflection across the y-a

xis, then a translation a units to the right and b units up translation a units to the right and b units up, then a rotation of 180 degrees counterclockwise about the origin, then a reflection across the y-axis
Mathematics
1 answer:
SpyIntel [72]3 years ago
7 0

Answer:

The point O can be (x,y)  or (-x,y) or (-x,-y) or ( x,-y) reason being that you haven't given point O lies in which quadrant.

Rotation through 90° counter clockwise

(x,y) = (-y,x)

(-x,y)=(-y,-x)

(-x,-y)=(y,-x)

(x,-y) =(y,x)

Then Reflection across X axis has taken place.

(-y,x) = (-y,-x)

(-y,-x)=(-y,x)

(y,-x)=(y,x)

(y,x)=(y,-x)

Then a reflection across the y-axis has taken place.

(-y,-x)=(y,-x)

(-y,x) = (y,x)

(y,x) =(-y,x)

(y,-x)=(-y,-x)

Then a translation a units to the right and b units up has taken place.

(y,-x)=(y+a,-x+b)

(y,x) =(y+a, x+b)

(-y,x) =(-y+a,x+b)

(-y,-x)=(-y+a,-x+b)

Then a rotation of 180 degrees counterclockwise about the origin has taken place.

(y+a,-x+b)=[-(y+a),-(-x+b)]

(y+a, x+b)=[- (y+a), -(x+b)]

(-y+a,x+b)=[-(-y+a),-(x+b)]

(-y+a,-x+b) =[-(-y+a),-(-x+b)]

Now Again a reflection across the Y axis has taken place.

[-(y+a),-(-x+b)]=[(y+a),-(-x+b)]

[-(y+a),-(x+b)]=[(y+a),-(-x+b)]

[-(-y+a),-(x+b)]=[(-y+a),-(x+b)]

[-(-y+a),-(-x+b)]=[(-y+a),-(-x+b)]

Totally depends on value of a and b on which quadrant these point lies.



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