Answer #1: the answer is the triangles new coordinates are A(4,3)B(-2,7)C(0,0)

Explanation: first I copied the triangle down on to a clear piece of paper the I rotated the triangle counter clockwise 90 degrees on the paper. Then i copied down the new coordinates on to another paper.

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Answer #2: The new points will be as follows: A'=(-1,2), B'=(-7,5), and C'=(-4,-2)

Explanation: I used Geometer's Sketchpad to solve this problem. I made a graph, plotted points, made point E the center of rotation, and rotated around the point by 90 degrees. Then I read the points off of the graph, and transferred them to this sheet.

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Answer#3: After the triangle is rotated 90 degrees about the origin the new coordinates are A(-2,1) B(-5,7) C(2,4).

Explanation: The rules for rotating a triangle 90 degrees about the origin is (x,y)--->(-x,y). Therefore, point A(2,1) rotated to (-2,1), point B (5,7) rotated to (-5,7) and point C(-2,4) rotated to (2,4).

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Answer #4: A : (-2, -1), B : (-8, 2), C : (-5, -5)

Explanation: first, i found the perpendiculars of each segment of the triangle,meaning it would give me what that segment would be 90 degrees counterclockwise on the graph. Then with that, i could find each one of the 3 vertices of the triangle by simply connecting the perpendiculars. with a triangle formed 90 degrees from the original using vertice A as the origin, i just moved the entire triangle down one and across left two spaces to make vertice A on the origin of the graph, and then i did the same thing again, so i used point A and i moved it down one space and over left two spaces to make is exactly opposite of the initial triangle, and so, that would give me the triangle's position 90 degrees counterclockwise from the initial triangle using the origin of the graph as the turning point. thanks!

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Answer #5: After the triangle is rotated 90 degrees (counterclockwise) about the origin, The new vertices will be: A'=(-1.5, 2.5) B'=(-8.5, 5) and C'=(-4.5, -2).

Explanation: First, I drew the triangle on a graph, with its vertices labeled A,B, and C corresponding to the picture shown above on the web page. Then, I constructed a line segment from vertex A of the triangle to the point of rotation, which in this case is the origin, and did the same with vertex B and C. Then, using my protractor, I measured 90 degrees (counterclockwise, which is the positive direction of measuring degrees) putting the line segment at zero degrees and marking 90 degrees counterclockwise. After that, I constructed another line segment from the origin to the place where I marked off the 90 degrees counterclockwise rotation.
Then, I used my compass to construct a 90 degree arc from the vertex of the triangle to the line segment that was 90 degrees counter clockwise away from the vertex. I marked the point where the arc and the line segment intersected as the image point A', and I repeated the same process for vertices B and C, marking the image points B' and C'. Finally, I drew the segments A'B', B'C', and A'C', making the triangle A'B'C', which is the image of triangle ABC rotated 90 degrees (counterclockwise).
The coordinates for vertex A' were (-1.5, 2.5)
The coordinates for vertex B' were (-8.5, 5)
The coordinates for vertex C' were (-4.5, -2)

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Answer#6: C (4,2), A (1,-2), B (7,-5)

Explanation: First I re-graphed the traingle on a separate peice of graph paper so that the new coordinates could fit. I decided to work with point C first. I drew a straight line from point C to the origin. Then i used the edge of a piece of paper for my compass because it is exactly 90 degrees. i placed one side of the paper along the line i just drew facing the 1st and 4th quadrent. Then i drew another line along the other side of the paper. The new coordinate for point C would lie somewhere along this line.
For point A i drew a straight line from A to the origin, i also did this with point B. Then with both points i followed the same type of procedure as i did with C. The edge of my compass paper would always face the right because i was going a positive 90 degrees not negative.
When i had all the lines drawn to show where the three points would lie along i took out a ruler. I measured how far each point was from the origin. Then, i took this distance and measured the same amount along the new lines that i had just drawn for each point.
Point C landed at (4,2) in the 1st quadrent. Point A was (1,-2) in quadrent 4. Point B was in quadrent 4 also, at (7,-5).

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