Question

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 1

D. Rotation 270 degrees clockwise.

Explanation:

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

`(x', y') = (r\cdot \cos (\theta - 270^(\circ)), r\cdot \sin (\theta -270^(\circ)))`

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

Where
`r` is the norm of original point, defined as:

`r =\sqrt{x^(2)+y^(2)}`,
`\forall \,x,y\in \mathbb{R}`

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

`\theta = \tan^(-1) (y)/(x)`

Now we proceed to prove the statement:

`L (x, y) = (-1, 9) \longrightarrow L'(x,y)`

Norm

`r = \sqrt{(-1)^(2)+9^(2)}`

`r = √(82)`

Direction

The point is located on 2nd quadrant, which means that
`90^(\circ) \leq \theta \leq 180^(\circ)`. Then:

`\theta = \tan^(-1)\left((9)/(-1) \right)`

`\theta = 96.340^(\circ)`

Rotation

`L'(x,y) = (√(82)\cdot \cos (96.340^(\circ)-270^(\circ)),√(82)\cdot \sin (96.340^(\circ)-270^(\circ)))`

`L'(x,y) = (-9, -1)`

`M(x,y) = (-8,8) \longrightarrow M'(x,y)`

Norm

`r = \sqrt{(-8)^(2)+8^(2)}`

`r = √(128)`

Direction

The point is located on 2nd quadrant, which means that
`90^(\circ) \leq \theta \leq 180^(\circ)`. Then:

`\theta = \tan^(-1)\left((8)/(-8) \right)`

`\theta = 135^(\circ)`

Rotation

`M'(x,y) = (√(128)\cdot \cos (135^(\circ)-270^(\circ)),√(128)\cdot \sin (135^(\circ)-270^(\circ)))`

`M'(x,y) = (-8, -8)`

`N(x,y) = (-3,5)\longrightarrow N'(x,y)`

Norm

`r = \sqrt{(-3)^(2)+5^(2)}`

`r =√(34)`

Direction

The point is located on 2nd quadrant, which means that
`90^(\circ) \leq \theta \leq 180^(\circ)`. Then:

`\theta = \tan^(-1)\left((5)/(-3) \right)`

`\theta = 120.964^(\circ)`

Rotation

`N'(x,y) = (√(34)\cdot \cos (120.964^(\circ)-270^(\circ)), √(34)\cdot \cos (120.964^(\circ)-270^(\circ)))`

`N'(x,y) = (-5, -3)`

Hence, the correct answer is D.