`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)`
