Question

A triangle is drawn on a coordinate plane. Dilate the figure with a magnitude of 3 in reference to the origin. What are the coordinates of the new vertices of A′B′C′ ? (1 point) Responses A′(1,−2) , B′(3,6) , and C′(9,0) upper A prime left parenthesis 1 comma negative 2 right parenthesis , upper B prime left parenthesis 3 comma 6 right parenthesis , and upper C prime left parenthesis 9 comma 0 right parenthesis A′(−6,−15) , B′(0,9) , and C′(18,−9) upper A prime left parenthesis negative 6 comma negative 15 right parenthesis , upper B prime left parenthesis 0 comma 9 right parenthesis , and upper C prime left parenthesis 18 comma negative 9 right parenthesis A′(−23,−53) , B′(0,1) , and C′(2,−1) upper A prime left parenthesis negative Start Fraction 2 over 3 End Fraction comma negative Start Fraction 5 over 3 End Fraction right parenthesis , upper B prime left parenthesis 0 comma 1 right parenthesis , and upper C prime left parenthesis 2 comma negative 1 right parenthesis A′(−5,−8) , B′(−3,0) , and C′(3,0)

Answer 1

To dilate a figure with a magnitude of 3 in reference to the origin, we multiply each coordinate of the vertices by 3.

Let's denote the original vertices of the triangle as follows:
- A(x₁, y₁)
- B(x₂, y₂)
- C(x₃, y₃)

The coordinates of the new vertices after the dilation will be:
- A′(3x₁, 3y₁)
- B′(3x₂, 3y₂)
- C′(3x₃, 3y₃)

However, we need the original coordinates of points A, B, and C in order to perform the dilation and find A′B′C′. Since the original coordinates are not given in your question, we cannot perform the calculation without that information.

If you provide the original coordinates of the vertices A, B, and C, I can help you find the coordinates of the new vertices A′, B′, and C′.