Tamika is trying to dilate a triangle drawn on a coordinate plane using a magnitude of 4 with reference to the origin. If the original triangle has vertices at coordinate points (12,0) , (5,0) , and (4,3) , what are the coordinate points of the new vertices of the dilated triangle?(1 point)Responses(18,0), (54,0), and (1,34)left parenthesis Start Fraction 1 over 8 End Fraction comma 0 right parenthesis , left parenthesis Start Fraction 5 over 4 End Fraction comma 0 right parenthesis , and left parenthesis 1 comma Start Fraction 3 over 4 End Fraction right parenthesis(2,0), (20,0), and (16,12)left parenthesis 2 comma 0 right parenthesis , left parenthesis 29 comma 0 right parenthesis , and left parenthesis 16 comma 12 right parenthesis(92,4), (9,4), and (8,7)left parenthesis Start Fraction 9 over 2 End Fraction comma 4 right parenthesis , left parenthesis 9 comma 4 right parenthesis , and left parenthesis 8 comma 7 right parenthesis(−72,−4), (1,−4), and (0,−1)

Answer 1

To dilate a triangle from the origin by a scale factor of 4, we multiply each coordinate of the triangle’s vertices by 4.

The original vertices are:

1. \( (12, 0) \)
2. \( (5, 0) \)
3. \( (4, 3) \)

Now, we will perform the dilation for each vertex:

1. For \( (12, 0) \): \[ (12 \times 4, 0 \times 4) = (48, 0) \]

2. For \( (5, 0) \): \[ (5 \times 4, 0 \times 4) = (20, 0) \]

3. For \( (4, 3) \): \[ (4 \times 4, 3 \times 4) = (16, 12) \]

So, the new vertices of the dilated triangle are \( (48, 0) \), \( (20, 0) \), and \( (16, 12) \).

Looking at the response options you provided, the correct answer corresponds to \( (20, 0) \), \( (48, 0) \), and \( (16, 12) \), which is not explicitly listed as given. Nonetheless, the vertices after dilation are \( (48, 0) \), \( (20, 0) \), and \( (16, 12) \).