Dilate line f by a scale factor of 2 with the center of dilation at the origin to create line f. Where are points A' and B' located after dilation, and how are lines fand f related? (A) The locations of A' and B' are A' (0,4) and B' (4,0); lines f and f are parallel.

(B) The locations of A' and B' are A' (0, 2) and B' (2, 0); lines f and f are the same line.

(c) The locations of A' and B' are A' (0, 2) and B' (4, ); lines f and f intersect at point A.

(D) The locations of A' and B' are A' (0,4) and B' (2, 0); lines f and f intersect at point B.

Option A

Step-by-step explanation:

We are given that

Scale factor=2

Center of dilation=(0,0)

Point A(0,2) and point B (2,0).

Slope of line f=

Substitute the values

Slope of line f=

Distance between  A and Origin (0,0) is given by

Using distance formula

OB=

Length of OA'=2OA=2(2)=4 units

Length of OB'=2(OB)=2(2)=4 units

x-intercept of line f' at x=4

y-intercept of line f' at y=4

Therefore, the points A' and B' are given by

(0,4) and (4,0)

Slope of line f'=

Slope of line f and f' are equal.Therefore, lines f and f' are parallel.

Option A is true.

x=6 and x=3

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< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)