After a rotation of 90° about the origin, the coordinates of the vertices of the image of a triangle are a'(6, 3), b'(–2, 1), and c'(1, 7). what are the coordinates of the vertices of the pre-image?

Assuming the original was rotated counterclockwise, and this only works for a rotation about the origin, a rotation of 90 degrees switches the x and y coordinate and depending on which part of the graph (quadrant) it is in the signs may change as well.  First, and the trickiest part is finding how the signs change. 

There are four quadrants, first, second third and fourth starting with the upper right than upper left than lower left then lower right.Since none of the veracities have a 0 in them we don't have to worry about the x or y axis, though it is worth realizing each axis is 90 degress from the others.  So something  at 2 on the x axis (2,0) moves 90 degrees counterclockwise would then be at 2 on the y axis (0,2) and moves one more time would be at (-2,0).

A' is in the first quadrant (upper right) so the previous quadrant 90 degrees clockwise is the fourth quadrant (lower right).  The signs of the coordinates then will go from (+,+) to (+,-).

Now that we have the signs we just switch the coordinates so (6,3) goes to (3,6) and the sign change makes it (3,-6).  Just follow this pattern with the other two points (-2,1), and (1,7) find the signs by figuring out the previous quadrant than switch the coordinates and apply the sign changes

The distance from point a to point c across the coordinate plane is aproxamatly 8.94 rounded to the nearest hundredth. you use a^2 + b^2 =c^2

Use a graph site then you will find your answer