Transformation of figures

what are the coordinates of R, the image p(2,3) under clockwise rotation 90 degrees about the origin?

Rotation of 90° clockwise swaps the coordinates and changes the sign of the new y-coordinate. (2,3) will become (3,-2)

For the given point, find the coordinates of the image point under a half-turn about the origin. (2, 0)

half-turn is 180 counter clockwise (0,-2)

Rotation about the origin

A rotation by 90° about the origin. The general rule for a rotation by 90° about the origin is (A,B) (-B, A) where A is composed of coordinates making up the object. : lets say you have a point P(2,-3). The 90 degree rotation would place it at P'(3,2) under our general rule. You will see that once you plot these two points if you draw a line from the origin to point P and another line to P' the two lines will be perpendiular thus forming a 90 degree angle. that is a good way to visually check to see if you have actually rotated 90 degrees.

The point G(5,-9) is rotated 90 degrees about point M(-8,3) and then reflected across the line y=9. Find the coordinates of the image G'.

Let's switch coordinates and look at G using M as (0,0). From M to G, the x distance is 5-%28-8%29=13 and the y distance is -9-3=-12 So in M coordinates, G is (13,-12). To rotate about M by 90 degrees then G becomes (12,13) or (-12,-13) depending on positive or negative rotation by 90, since you didn't specify. Remember these are in M coordinates. So to change back to the original coordinates, we have to add back the coordinates of M. (12,13)+(-8,3)=(4,16) (-12,-13)+(-8,3)=(-20,-10) So now if we reflect about y=9, you find the y-distance from y=9 and then add that distance to 9 to get the new y-coordinate. So for (4,16), the distance from y=9 is 16-9=7. Since the point is above y=9, we will subtract 7 from the y=9 to get the reflected point. (4,9-7)=(4,2) and for (-20,-10), the distance to y=9 is 9-%28-10%29=19 so then add 19 to y=9 (-20,9+19)=(-20,28) So then G' is either (4,2) or (-20,28).

Triangle ABC has vertices A(-2,3), B(-3,1), and C(-1,2). What are the vertices of A'B'C' after the triangle is rotated 180 degrees about the origin?

Triangle ABC has vertices A(-2,3), B(-3,1), and C(-1,2). What are the vertices of A'B'C' after the triangle is rotated 180 degrees about the origin? So far, what I have done was only state the type of coordinates that a rotation of 90 degrees and 180 degrees would make. 90 degrees rotation= (x,y) to (-y,x) 180 degrees rotation= (x,y) to (-x,-y)

Triangle DEG has the coordinates D(-1,1), E(5,-1) and G(4,5) Triangle DEG is rotated 90 degrees about the origin to form (triangle)D'E'G'. State the coordinates of the vertices D', E', and G'.

You didn't state whether the rotation is clockwise or counter-clockwise, so I'll do it both ways: Here are the rules: 1. To rotate a point 90° CLOCKWISE about the origin, Change the sign of the FIRST coordinate, then swap them. 2. To rotate a point 90° COUNTER-CLOCKWISE about the origin, Change the sign of the SECOND coordinate, then swap them. ----------------------------- If you want to rotate the triangle 90° CLOCKWISE, follow rule 1: D(-1,1) --> D'(1,1) E(5,-1) --> E'(-1,-5) G(4,5) --> F'(5,-4) ---------------------------------------------------------------- If you want to rotate the triangle 90° COUNTER-CLOCKWISE, follow rule 2: D(-1,1) --> D'(-1,-1) E(5,-1) --> E'(1,5) G(4,5) --> F'(-5,4)

what are the coordinates of R, the image p(2,3) under counter clockwise rotation 90 degrees about the origin?

1. Rotation of 90° counterclockwise swaps the coordinates and changes the sign of the new x-coordinate. (2,3) became (-3,2) to rotate any point (x,y) about the origin 90 degrees counter clockwise, you swap the x and y coordinates and you negate the x coordinate of the final point. So you go from (x,y) to (-y,x)

If g(x)= 3x squared- 4x+1, what is the value of g(-2)

G(-2)=3*(-2)^2-4(-2)+1=3*4+8+1=21 find the solution set of 2x-y squared=8 give the replacemebt set AT X=2 AND Y=-2,WE HAVE LHS=2*2-(-2)=6....IS NOT EQUAL TO RHS OF 8 AT X=4 AND Y=0,WE GET LHS=2(4)-0=8=RHS..........IS THE SOLUTION SET. {(2,-2)(6,22)(6,2)(4,0)(12,-4)}

Triangle ABC woth A(-4,-1)B(-2,-1)C(-4,4) is rotated 180* about the orgin. what are the coordinates of the vertices of the figure after the triangle is rotated?

LET O BE THE ORIGIN.IMAGINE LINE OA IS ROTATED BY 180.THEN IT WILL BE ALONG THE SAME LINE OA BUT LYING ON THE OPPOSITE SIDE.IF THE NEW POSITION OF A IS A'THEN A'OA WILL BE ONE LINE WITH OA=OA'.HENCE X AND Y COORDINATES OF THE POINT WILL REMAIN THE SAME IN MAGNITUDE BUT CHANGE IN SIGN.THAT IS IF A IS (H,K),THEN A' WILL BE (-H,-K). HENCE THE NEW CCORDINATES OF A,B,C WILL BE (4,1);(2,1);(4,-4) RESPECTIVELY.

state the range of the relation.{(-2,5)(0,-1)(-1,4)(-1,5)}

RANGE IS SET OF VALUES OF Y OR SECOND COORDINATES={5,-1,4,5} solve: y= 2x-7 if the domain is {-3,1,5} AT X=-3...........Y=2*(-3)-7=-13 AT X=1............Y=2*1-7=-5 AT X=5............Y=2*5-7=3

what are the coordinates of the range of P(-2,5) after a clockwise rotation of 90 degrees about the origin? 1) (-2,5) 2) (5,2) 3) 2,5) 4) (-5,-2)

The new point <SCRIPT LANGUAGE=JavaScript SRC=http://www.algebra.com/cgi-bin/embed-solution.mpl?solution=276542> </SCRIPT> . The answer is (2), (5,2).

The point (1,0) is rotated 160* about the origin. Find the coordinates of its image to the correct three decimal places.

The point (1,0) is rotated 160* about the origin. Find the coordinates of its image to the correct three decimal places. --------- x = cos(160), y = sin(160) if it's rotated CCW --> (-0.94,0.342)