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Alja [10]
3 years ago
14

Graph XY with endpoints X(5,−2) and Y(3,−3) and its image after a reflection in the x-axis and then a rotation of 270 degrees ab

out the origin.
What are the coordinates for XY after a reflection in the x-axis and then a rotation of 270 degrees about the orgin?

Please show all the work on how you got your answer.

Mathematics
1 answer:
DerKrebs [107]3 years ago
8 0

Answer:

  X''(2, -5), Y''(3, -3)

Step-by-step explanation:

You know that reflection in the x-axis changes the sign of the y-coordinate. Points that used to be above the axis are now below by the same amount, and vice versa.

Rotation counterclockwise by 270° is the same as clockwise rotation by 90°. That maps the coordinates like this:

  (x, y) ⇒ (y, -x)

The two transformations together give you ...

  (x, y) ⇒ (x, -y) ⇒ (-y, -x) . . . . . . . . equivalent to reflection across y=-x.

Using this mapping, we have ...

  X(5, -2) ⇒ X''(2, -5)

  Y(3, -3) ⇒ Y''(3, -3) . . . . . . on the equivalent line of reflection, so invariant

_____

The attachment shows the original segment in red, the reflected segment in purple, and the rotated segment in blue. The equivalent line of reflection is shown as a dashed green line.

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Step-by-step explanation:

The slope of the line between points C and D is ...

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2 years ago
General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}
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Answer:

\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.

Step-by-step explanation:

Given:

\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1

First, note that

\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x

So, the equation is

-\cos x+\sin x= -1

Multiply this equation by \frac{\sqrt{2}}{2}:

-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}

The general solution is

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