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On a coordinate plane, triangle A B C has points (negative 9, 3), (negative 9, 6), (0, 3) and triangle A double-prime B double-prime C double-prime has points (3, negative 1), (3, negative 2), and (0, negative 1). Which transformations could be performed to show that △ABC is similar to △A"B"C"? a reflection over the x-axis, then a dilation by a scale factor of 3 a reflection over the x-axis, then a dilation by a scale factor of One-third a 180° rotation about the origin, then a dilation by a scale factor of 3 a 180° rotation about the origin, then a dilation by a scale factor of One-third

On A Coordinate Plane Triangle A B C Has Points Negative 9 3 Negative 9 6 0 3 And Triangle A Doubleprime B Doubleprime C Doubleprime Has Points 3 Negative 1 3 N class=

Sagot :

Answer:

Option (4)

Step-by-step explanation:

Vertices of ΔABC are,

A → (-9, 3)

B → (-9, 6)

C → (0, 3)

If the given triangle is rotated 180° about the origin image triangle will have the vertices as,

Rule for the rotation,

(x, y) → (-x, -y)

A' → (9, -3)

B' → (9, -6)

C' → (0, -3)

If a point (x, y) is dilated by a scale factor of 'k',

Rule for the dilation,

(x, y) → (kx, ky)

Since, ΔA"B"C" is smaller than ΔABC,

Therefore, dilating ΔA'B'C' by a scale factor = [tex]\frac{1}{3}[/tex]

Image triangle ΔA"B"C" will be,

A" → (3, -1)

B" → (3, -2)

C" → (0, -1)

Therefore, Option (4) is the correct option.

Answer:

the guy above me is wrong the correct option is the second option

Step-by-step explanation:

i just got it wrong & it showed me the answer

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