You haven't provided the coordinates of the original trapezium, therefore, I can only help with the concept.
From the givens, we can know that the rotation was 90° counter-clockwise about the origin.
The rule for this rotation is:
(x , y) ..............> (-y , x)
This means that to get the points of the rotated image, negate the y-value then exchange it with the x-value
Examples:
(1 , 3) rotated 90° counter-clockwise about the origin will give (-3 , 1)
(-2 , 4) rotated 90° counter-clockwise about the origin will give (-4 , -2)
(1 , -2) rotated 90° counter-clockwise about the origin will give (2 , 1)
(-3 , -5) rotated 90° counter-clockwise about the origin will give (5 , -3)
Hope this helps :)
Answer:
F'(-5,-7) and G'(-3,-4).
Step-by-step explanation:
Consider the below figure attached with this question.
From the below figure it is clear that the vertices of trapezoid are E(-4,8), F(-7,5), G(-4,3) and H(-2,5).
If a figure is rotated 90° about the origin in the counterclockwise direction, then the rule of rotation is
[tex](x,y)\rightarrow (-y,x)[/tex]
We need to find the coordinates of the endpoints of the side congruent to side FG. It means we have to find F' and G'.
Using the above rule we get
[tex]F(-7,5)\rightarrow F'(-5,-7)[/tex]
[tex]G(-4,3)\rightarrow G'(-3,-4)[/tex]
Therefore, the endpoints of the side congruent to side FG are F'(-5,-7) and G'(-3,-4).
