Answer:A
Step-by-step explanation:
You have not provided the diagram/coordinates for point Q, therefore, I cannot provide an exact answer.
However, I can help you with the concept.
When rotating a point 90° counter clock-wise, the following happens:
coordinates of the original point: (x,y)
coordinates of the image point: (-y,x)
Examples:
point (2,5) when rotated 90° counter clock-wise, the coordinates of the image would be (-5,2)
point (1,9) when rotated 90° counter clock-wise, the coordinates of the image would be (-9,1)
point (7,4) when rotated 90° counter clock-wise, the coordinates of the image would be (-4,7)
Therefore, for the given point Q, all you have to do to get the coordinates of the image is apply the transformation:
(x,y) .............> are changed into.............> (-y,x)
Hope this helps :)
Assume your line starts at zero, your first point is (-3,5) meaning your have a slope of 5/-3
[ f(x) = 5/-3x + b]
You could just use 3.14 or 22/7
The answer is B, because the numbers in the equation are 2, and -3x. The signs have to stay with the numbers they are with. So if you move the 3 to the front you have to keep the negative with it.
