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Which of the following represents a rotation of triangle XYZ, which has vertices (-4,7), Y(6,2), and Z (3,-8) about the origin by 90 degrees? HELP PLS options: A: X (-7,-4) Y(6,-2) Z(-8,3) B: X(7,-4) Y(-2,6) Z (3,-8) C: X (-7,-4) Y(-2,6) Z (8,3) D: X(7,-4) Y (-2,6) Z (-3,8)

User James Gan
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1 Answer

4 votes

Answer:

The best option is B: X' = (-7,-4), Y' = (-2,6), Z'=(8, 3).

Explanation:

Each vertex can be represented as a vector with regard to origin.


\vec X = -4\cdot i + 7\cdot j,
\vec Y = 6\cdot i + 2\cdot j and
\vec Z = 3\cdot i -8\cdot j.

The magnitudes and directions of each vector are, respectively:

X:


\|\vec X\| = \sqrt{(-4)^(2)+7^(2)}


\|\vec X\| \approx 8.063


\theta_(X) = \tan^(-1)\left((7)/(-4) \right)


\theta_(X) \approx 119.744^(\circ)

Y:


\|\vec Y\| = \sqrt{6^(2)+2^(2)}


\|\vec Y\| \approx 6.325


\theta_(Y) = \tan^(-1)\left((2)/(6) \right)


\theta_(Y) \approx 18.435^(\circ)

Z:


\|\vec Z\| = \sqrt{3^(2)+(-8)^(2)}


\|\vec Z\| \approx 8.544


\theta_(Z) = \tan^(-1)\left((-8)/(3) \right)


\theta_(Z) \approx 290.556^(\circ)

Now, the rotation consist is changing the direction of each vector in
\pm 90^(\circ), which means the existence of two solutions. That is:


\vec p = r \cdot [\cos (\theta \pm 90^(\circ))\cdot i + \sin (\theta \pm 90^(\circ))\cdot j]

Where
r and
\theta are the magnitude and the original angle of the vector.

Solution I (
+90^(\circ))


\vec p_(X) = 8.063\cdot [\cos (119.744^(\circ)+90^(\circ))\cdot i + \sin (119.744^(\circ)+90^(\circ))\cdot j]


\vec p_(X) = -7\cdot i -4\cdot j


\vec p_(Y) = 6.325\cdot [\cos(18.435^(\circ)+90^(\circ))\cdot i+\sin(18.435^(\circ)+90^(\circ))\cdot j]


\vec p_(Y) = -2\cdot i +6\cdot j


\vec p_(Z) = 8.544\cdot [\cos(290^(\circ)+90^(\circ))\cdot i +\sin(290^(\circ)+90^(\circ))\cdot j]


\vec p_(Z) = 8.029\cdot i +2.922\cdot j

Solution II (
-90^(\circ))


\vec p_(X) = 8.063\cdot [\cos (119.744^(\circ)-90^(\circ))\cdot i + \sin (119.744^(\circ)-90^(\circ))\cdot j]


\vec p_(X) = 7\cdot i +4\cdot j


\vec p_(Y) = 6.325\cdot [\cos(18.435^(\circ)-90^(\circ))\cdot i+\sin(18.435^(\circ)-90^(\circ))\cdot j]


\vec p_(Y) = 2\cdot i -6\cdot j


\vec p_(Z) = 8.544\cdot [\cos(290^(\circ)-90^(\circ))\cdot i +\sin(290^(\circ)-90^(\circ))\cdot j]


\vec p_(Z) = -8.029\cdot i -2.922\cdot j

The rotated vertices are: i) X' = (-7,-4), Y' = (-2,6), Z'=(8.029, 2.922) or ii) X' = (7,4), Y' = (2,-6), Z' = (-8.029, -2.922). The best option is B: X' = (-7,-4), Y' = (-2,6), Z'=(8, 3).

User RKS
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