A triangle contains vertices A (-3, 2), B (1, 2), and C (-4, -2).

If ∆ABC is rotated 90° counterclockwise around the origin, what are the coordinates of the transformed figure ∆A’B’C’ ?


A ’(-2, 3), B ’(-2, 1), C ’(2, -4)
A ’(-2, 3), B ’(-2, 1), C ’(2, -4)

A ’(-3, -2), B ’(1, -2), C ’(-4, 2)
A, ’(-3, -2), , B, ’(1, -2), , C, ’(-4, 2)

A ’(-2, -3), B ’(-2, 1), C ’(2, -4)
A, ’(-2, -3), , B, ’(-2, 1), , C, ’(2, -4)

A ’(-3, 2), B ’(1, 2), C ’(4, -2)
A, ’(-3, 2), , B, ’(1, 2), , C, ’(4, -2)

When a shape is rotated, it must be rotated through a point

The coordinates of the transformed figure are: A ’(-2, 3), B ’(-2, 1), C ’(2, -4)

The coordinates of the triangle is given as:

[tex]\mathbf{A = (-3,2)}[/tex]

[tex]\mathbf{B = (1,2)}[/tex]

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

The rule of 90 degrees counterclockwise rotation is:

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

So, we have:

[tex]\mathbf{A(-3,2) \to (-2,-3)}[/tex]

[tex]\mathbf{B(1,2) \to (-2,1)}[/tex]

[tex]\mathbf{C(-4,-2) \to (2,-4)}[/tex]

Hence, the coordinates of the transformed figure are:

A ’(-2, 3), B ’(-2, 1), C ’(2, -4)

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