What happens to a vertical line if you rotate it 90 degrees clockwise?

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The point (-3, ii) is rotated ninety° clockwise around the origin to point [#permalink] 21 Mar 2019, 23:13

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The betoken (-3, 2) is rotated 90° clockwise around the origin to signal B. Point B is then reflected over the line ten = y to point C. What are the coordinates of C?

(A) (-three, -2)
(B) (-2, -3)
(C) (two, -3)
(D) (2, 3)
(E) (3, 2)

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Re: The point (-3, 2) is rotated xc° clockwise around the origin to point [#permalink] 22 Mar 2019, 00:53

IMO D

Afterwards 90 degree clockwise rotation, it goes to fiest quadrant , the point become (iii,two )

and the refelection around y=x ia given by changing the (x,y ) coordinate

hence answer is (2,iii )

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The point (-3, two) is rotated 90° clockwise around the origin to point [#permalink] 22 Mar 2019, 01:27

Bunuel wrote:

The point (-three, 2) is rotated 90° clockwise effectually the origin to betoken B. Bespeak B is so reflected over the line x = y to betoken C. What are the coordinates of C?

(A) (-iii, -2)
(B) (-2, -3)
(C) (2, -3)
(D) (2, 3)
(E) (three, 2)

clockwise direction motility around origin will event (-iii,two) to go ( 2,3)
the pt C coordinates x=y is 1st quadrant ; later which the coordinate gets reflected so indicate becomes iii,2
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Re: The point (-iii, ii) is rotated 90° clockwise around the origin to point [#permalink] 13 Jul 2020, 21:19

Bunuel wrote:

The bespeak (-three, ii) is rotated 90° clockwise around the origin to point B. Signal B is then reflected over the line x = y to bespeak C. What are the coordinates of C?

(A) (-iii, -2)
(B) (-2, -3)
(C) (2, -3)
(D) (two, 3)
(E) (3, 2)

Given: The signal (-3, two) is rotated 90° clockwise around the origin to point B. Point B is then reflected over the line ten = y to signal C.
Asked: What are the coordinates of C?

Attachment:

Screenshot 2020-07-14 at 10.49.02 AM.png [ 26.thirteen KiB | Viewed 12030 times ]

IMO E
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The betoken (-3, two) is rotated 90° clockwise around the origin to betoken [#permalink] 24 Mar 2021, 18:30

Reflection 1:

The manner I understood 90 degree rotations was past the "tipping the rectangle" method.

It sounds goofy, merely:

Start with Original Signal (-3, 2)

Since we are rotating 90 degrees most the Origin (0 , 0) ——> brand a rectangle with these 2 points

Label (-3 , ii) as point A and Origin as Signal O

If we bring together the horizontal and vertical lines to the Y and X centrality, respectively, we end up with a rectangle that is three units along the Negative X Axis and 2 units along the Positive Y Centrality

Now imagine lifting the rectangle upwards and pushing it from quadrant two into quadrant 1 - i.e, a ninety degree rotation clockwise about the Origin

This would give united states of america a rectangle of 2 units along the positive X axis and 3 units along positive Y centrality. Point A would now motion to the upper right corner of this new rectangle that nosotros pushed over ———> this volition be indicate (ii , three)

You can ostend visually past:

-connecting original signal A at (-3 ,2) with the Origin

-and connecting new image indicate A at (2 , iii)

these 2 points and the origin will create a 90 degree angle nearly the Origin (both lines are diagonals of their respective rectangles)

Or, the caption is besides long winded, you lot can follow the Dominion:

For 90 degree rotations clockwise about the Origin: (X , Y) ———>becomes prototype signal of (Y , -X)

(-iii ,two) becomes (2, iii)

So the starting time rotation gives us point (ii , iii)

Reflection 2:

Line Y = Ten is the Line that passes through the origin and creates a 45 degree bending with the X centrality

The original point of (2 , three) and the Image Signal will always be equidistant from the Mirror Line over which the original betoken is reflected.

To visualize it in this problem:

(2 ,2) and (3 , 3) are both on line Y = X

Point (2 , 3) will be +1 unit above (ii ,2) and + 1 unit of measurement to the left of (iii , 3)

This creates a right triangle with sides of 1 and ane

To detect the image point reflected over Y = Ten, nosotros brand the reverse moves from these points (reversed) on Line Y = X:

from (2 , ii) we move 1 unit to the correct: Ten coordinate = iii

From (iii , 3) we move ane unit downwards: Y coordinate = 2

Last Point is (3 , 2)

Or the rule:

Reflecting point (X , Y) over the line given by Y = X ————> image point volition be (Y , Ten)

In other words, just rearrange the coordinates

Respond Due east

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Re: The signal (-3, ii) is rotated 90° clockwise effectually the origin to point [#permalink] 24 Mar 2021, 23:nineteen

Bunuel wrote:

The signal (-3, 2) is rotated 90° clockwise around the origin to bespeak B. Point B is then reflected over the line 10 = y to point C. What are the coordinates of C?

(A) (-3, -2)
(B) (-2, -iii)
(C) (2, -3)
(D) (two, iii)
(E) (3, two)

We have discussed this concept in detail hither:

https://www.veritasprep.com/blog/2013/0 ... ry-function-i/

https://www.veritasprep.com/web log/2013/0 ... y-function-ii/

https://www.veritasprep.com/blog/2013/0 ... -function-iii/

When the point (-three, two) is rotated clockwise xc degrees, the betoken lies in outset quadrant and co-ordinates of x and y flip. So you get the point (2, 3).

When you reflect (two, iii) over the line x = y, this is like having a foursquare with one vertex at (2, 3), one at (2, ii), one at (3, 3) so the fourth vertex will be at (3, 2).

Answer (E)
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