Which transformation is shown by the coordinates below? L (-1, 9) M (-8, 8) N (-3, 5) L '(- 9, -1) M' (- 8, -8) N '(- 5, -3) A. Reflection over the x-axis / B. Translation 8 units left and 8 units down / C. Rotation 90 degrees clockwise / D. Rotation 270 degrees clockwise

Answer:

D. Rotation 270 degrees clockwise.

Step-by-step explanation:

The transformation used in the coordinates below was a rotation 270 degrees clockwise, whose definition is:

[tex](x', y') = (r\cdot \cos (\theta - 270^{\circ}), r\cdot \sin (\theta -270^{\circ}))[/tex]

Please note that clockwise rotation is represented by the minus sign.

Where [tex]r[/tex] is the norm of original point, defined as:

[tex]r =\sqrt{x^{2}+y^{2}}[/tex], [tex]\forall \,x,y\in \mathbb{R}[/tex]

And [tex]\theta[/tex] is the direction of the point centered at origin and with respect to +x semiaxis, measured in sexagesimal degrees:

[tex]\theta = \tan^{-1} \frac{y}{x}[/tex]

Now we proceed to prove the statement:

[tex]L (x, y) = (-1, 9) \longrightarrow L'(x,y)[/tex]

Norm

[tex]r = \sqrt{(-1)^{2}+9^{2}}[/tex]

[tex]r = \sqrt{82}[/tex]

Direction

The point is located on 2nd quadrant, which means that [tex]90^{\circ} \leq \theta \leq 180^{\circ}[/tex]. Then:

[tex]\theta = \tan^{-1}\left(\frac{9}{-1} \right)[/tex]

[tex]\theta = 96.340^{\circ}[/tex]

Rotation

[tex]L'(x,y) = (\sqrt{82}\cdot \cos (96.340^{\circ}-270^{\circ}),\sqrt{82}\cdot \sin (96.340^{\circ}-270^{\circ}))[/tex]

[tex]L'(x,y) = (-9, -1)[/tex]

[tex]M(x,y) = (-8,8) \longrightarrow M'(x,y)[/tex]

Norm

[tex]r = \sqrt{(-8)^{2}+8^{2}}[/tex]

[tex]r = \sqrt{128}[/tex]

Direction

The point is located on 2nd quadrant, which means that [tex]90^{\circ} \leq \theta \leq 180^{\circ}[/tex]. Then:

[tex]\theta = \tan^{-1}\left(\frac{8}{-8} \right)[/tex]

[tex]\theta = 135^{\circ}[/tex]

Rotation

[tex]M'(x,y) = (\sqrt{128}\cdot \cos (135^{\circ}-270^{\circ}),\sqrt{128}\cdot \sin (135^{\circ}-270^{\circ}))[/tex]

[tex]M'(x,y) = (-8, -8)[/tex]

[tex]N(x,y) = (-3,5)\longrightarrow N'(x,y)[/tex]

Norm

[tex]r = \sqrt{(-3)^{2}+5^{2}}[/tex]

[tex]r =\sqrt{34}[/tex]

Direction

The point is located on 2nd quadrant, which means that [tex]90^{\circ} \leq \theta \leq 180^{\circ}[/tex]. Then:

[tex]\theta = \tan^{-1}\left(\frac{5}{-3} \right)[/tex]

[tex]\theta = 120.964^{\circ}[/tex]

Rotation

[tex]N'(x,y) = (\sqrt{34}\cdot \cos (120.964^{\circ}-270^{\circ}), \sqrt{34}\cdot \cos (120.964^{\circ}-270^{\circ}))[/tex]

[tex]N'(x,y) = (-5, -3)[/tex]

Hence, the correct answer is D.

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