What is the result of rotating the triangle 90° counterclockwise around the point (2, -1)?
the first one is the one that is rotated

(0, 2) -> (-1, -3)

(1, 5) -> (-4, -2)

(3, 1) -> (0, 0)

Step-by-step explanation:

The general rule for a rotation 90° counterclockwise around the origin if you have a point (x, y) -> (-y, x). But since this isnt around the origin. You need to do a simple step before applying the rule.

To each point you want to rotate you need to substract the point that is the center of the rotation.

The points of the triangle are:

(0, 2)

(1, 5)

(3, 1)

Now you have to substract (2, -1) to each of them

(0, 2) - (2, -1) = (-2, 3)

(1, 5) - (2, -1) = (-1, 6)

(3, 1) - (2, -1) = (1, 2)

And with the new points you apply the rule for rotation (x,y) -> (-y,x)

(-2, 3) -> (-3, -2)

(-1, 6) -> (-6, -1)

(1, 2) -> (-2, 1)

And now you add the previously substracted point (center of rotation).

(-3, -2) + (2, -1) = (-1, -3)

(-6, -1) + (2, -1) = (-4, -2)

(-2, 1) + (2, -1 = (0, 0)

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.