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2. Given [tex]\triangle ABC[/tex] with vertices [tex]A(-3, -4)[/tex], [tex]B(0, -4)[/tex], and [tex]C(0, 1)[/tex], find the coordinates of [tex]\triangle A'B'C'[/tex], the image of [tex]\triangle ABC[/tex] with a scale factor of 2.

Sagot :

Sure, let's find the new coordinates of [tex]\(\triangle A^{\prime}B^{\prime}C^{\prime}\)[/tex] after scaling [tex]\(\triangle ABC\)[/tex] by a factor of 2.

1. Identify the original coordinates of vertices [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]:
- [tex]\(A(-3, -4)\)[/tex]
- [tex]\(B(0, -4)\)[/tex]
- [tex]\(C(0, 1)\)[/tex]

2. Apply the scale factor to each coordinate. The scale factor given is 2. This means we will multiply each coordinate of the vertices by 2.

3. Calculate the new coordinates for each vertex:

For [tex]\(A\)[/tex]:
[tex]\[ A^{\prime} = (2 \times -3, 2 \times -4) = (-6, -8) \][/tex]

For [tex]\(B\)[/tex]:
[tex]\[ B^{\prime} = (2 \times 0, 2 \times -4) = (0, -8) \][/tex]

For [tex]\(C\)[/tex]:
[tex]\[ C^{\prime} = (2 \times 0, 2 \times 1) = (0, 2) \][/tex]

4. Write the new coordinates of [tex]\(\triangle A^{\prime}B^{\prime}C^{\prime}\)[/tex]:
- [tex]\(A^{\prime} = (-6, -8)\)[/tex]
- [tex]\(B^{\prime} = (0, -8)\)[/tex]
- [tex]\(C^{\prime} = (0, 2)\)[/tex]

Therefore, the coordinates of [tex]\(\triangle A^{\prime}B^{\prime}C^{\prime}\)[/tex] after scaling by a factor of 2 are [tex]\((-6, -8)\)[/tex], [tex]\((0, -8)\)[/tex], and [tex]\((0, 2)\)[/tex].