What are the vertices of the image r(90°, O) (△BCD)? A. B'(−3, −3), C'(−1, 4), D'(2, 0)
B. B'(3, 3), C'(1, −4), D'(−2, 0)
C. B'(3, −3), C'(1, 4), D'(0, 2)
D. B'(−3, −3), C'(−1, −4), D'(0, −2)

The correct option is;

A. B'(-3, -3), C'(-1, 4), D'(2, 0)

Step-by-step explanation:

The given coordinates of the vertices triangle ΔBCD, are;

B(-3, 3), C(4, 1), D(0, -2)

The transformation to be performed is r(90°, 0) (ΔBCD) which is a 90 degrees counterclockwise rotation about the origin

The rule for transformation of a geometric figure by rotation of 90 degrees counterclockwise is given as follows;

The coordinates of the preimage (x, y), the coordinates of the image  (-y, x)

Therefore, we have;

The preimage B(-3, 3), the coordinates of the image after 90 degrees counterclockwise rotation is B'(-3, -3)

Similarly, we have;

The preimage C(4, 1), the coordinates of the image after 90 degrees counterclockwise rotation is C'(-1, 4)

For the point D, we have;

The preimage D(0, -2), the coordinates of the image after 90 degrees counterclockwise rotation is D'(2, 0)

Therefore, the correct option is B'(-3, -3), C'(-1, 4), D'(2, 0).

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )