Giúp mình với ạ!!! Thankssssss

`a)`

`T_{vec{v}} (M) = M’ (2 + 3; – 4 + 0)`

`=> M’ (5; -4)`

`b)` 

Vì `T_{vec{v}} (Δ) = Δ’`

`=> Δ’` có dạng: `-2x + 3y + c = 0`

Lấy điểm `A (1; 0) ∈ Δ`

`=> T_{vec{v}} (A) = A’ (1 + 2; -4 + 0)`

`=> A’ = (3; -4)`

`=> A’ ∈ Δ’`

`=> (-2).3 + 3.(-4) + c = 0`

`<=> c = 18`

`=> Δ’: -2x + 3y + 18 = 0`

`c)` Ta có

`(C)` \(\left\{ \begin{array}{l}I(2; -3)\\R = \sqrt{2^2 + (-3)^2 + 1} = \sqrt{14}\end{array} \right.\)

Gọi `T_{vec{v}} (I) = I’`

`=> I’ (2 + 2; -4 + 3) = (4; -1)`

`=> (C’)` có: \(\left\{ \begin{array}{l}I(4; -1)\\R = \sqrt{14}\end{array} \right.\)

`=> (C’): (x – 4)² + (y + 1)² = 14`

`d)` 

Ta có:

`T_{vec{v}} (A) = A’`

`=> A (- 1 – 2; 8 + 4) = (-3; 12)`