25 points triangle jkl is transformed by performing a 90 degree clockwise rotation about the origin and then a reflection over the y-axis, creating triangle j’’k’’l’’. which transformation will map j’’k’’l’’ back to jkl? a reflection over the y-axis and then a 90 degree clockwise rotation about the origin a reflection over the x-axis and then a 90 degree counterclockwise rotation about the origin a reflection over the x-axis and then a 90 degree clockwise rotation about the origin a reflection over the x-axis and then a reflection over the y-axis

The correct option is: Option: C.

A reflection over the x-axis and then a 90 degree clockwise rotation about the origin

Let us suppose the vertices of the original triangle JKL as:

J(0,0), K(1,0) and L(0,1)

Hence,when the triangle is rotated 90 degree clockwise about the origin, hence the vertices of the transformation are transformed by the rule:

(x,y) → (y,-x)

J(0,0) → J'(0,0)

K(1,0) → K'(0,-1)

L(0,1) → L'(1,0)

Now when this transformed figure is reflected over the y-axis, then the rule of transformed figure are:

(x,y) → (-x,y)

Hence,

J'(0,0) → J"(0,0)

K'(0,-1) → K"(0,-1)

L'(1,0) → L"(-1,0)

Hence, the transformation is:

J(0,0) → J''(0,0)

K(1,0) → K''(0,-1)

L(0,1) → L''(-1,0)

Hence, the transformation that map J"K"L" back to JKL are:

A reflection over the x-axis and then a 90 degree clockwise rotation about the origin.

( Since, on reflecting over the x-axis, the rule of transformation is:

(x,y) → (x,-y)

J"(0,0) → J'(0,0)

K"(0,-1) → K'(0,1)

L"(-1,0) → L'(-1,0)

When the transformed image is rotated 90 degree clockwise the rule of rotation is:

(x,y) → (y,-x)

Hence,

J'(0,0) → J(0,0)

K'(0,1) → K(1,0)

L'(-1,0) → L(0,1) )

congruent segments

step-by-step explanation: