Suppose the parallelogram JKLM has vertices: J (2, 1)

K (7, 1)

L (6, −3)

M (1, −3)

Write each transformation in function notation and find the new coordinates of each point.

1. J' if the parallelogram is rotated counterclockwise 270° about the origin?

2. K' if the parallelogram is rotated counterclockwise 180° about the origin?

3. L' if the parallelogram is rotated counterclockwise 90° about the origin?

4. M' if the parallelogram is rotated counterclockwise 90° about the origin?

I NEED HELP

1. For a rotation of 270° counterclockwise about the origin the coordinates of J' is (1, -2)

2. For a rotation of 180° counterclockwise about the origin the coordinates of K' is (-7, -1)

3. For a rotation of 90° counterclockwise about the origin the coordinates of L' is (3, 6)

4. For a rotation of 90° counterclockwise about the origin the coordinates of M' is (3, 1)

Step-by-step explanation:

The coordinates of the vertices of the parallelogram are given as follows;

J(2, 1), K(7, 1), L(6, -3), M(1, -3)

1. We have, for a rotation of 270° counterclockwise about the origin we have

f(x, y) = (y, -x)

Therefore, we have for the vertices;

f(J(2, 1)) = J'(1, -2)

Therefore, we have the coordinates of J' as (1, -2);

2. We have, for a rotation of 180° counterclockwise about the origin we have

f(x, y) = (-x, -y)

Therefore, we have for the vertex point K;

f(K(7, 1)) = K'(-7, -1)

3. We have, for a rotation of 90° counterclockwise about the origin we have

f(x, y) = (-y, x)

Therefore, we have for the vertex point L;

f(L(6, -3)) = L'(3, 6)

4. We have, for a rotation of 90° counterclockwise about the origin we have

f(x, y) = (-y, x)

Therefore, we have for the vertex point M;

f(M(1, -3)) = M'(3, 1)

hello from mrbilldoesmath!

answer:   x^2 + y^2 = 1024

discussion:

the equation of a circle with center (h,k) and radius "r" is

(x-h)^2 + (y-k)^2 = r^2.

in our case

(x-0)^2 + (y-0) ^2 = 32^2 = 1024

you,

mrb

for the free

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