Which of the following represents the translation of A(1,−2) along the vector <−3, 2> and then the vector <0, −4>? A (1, −2) → A ′(−2, 0) → A ″(−2, −4)
A (1, −2) → A ′(−2, 0) → A ″(−6, 0)
A (1, −2) → A ′(−3, −4) → A ″(0, −16)
A (1, −2) → A ′(3, −5) → A ″(3, −9)

what are the answear choices to the question.

step-by-step explanation:

its an acute right triangle

step-by-step explanation:

< 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)