What are the coordinates of the vertices of D(3, A) (△ABC) for A(0, 4), B(–2, 5), and C(3, 7)?

Question

Answers:

1. A. A′(0, 12), B′(−6, 15), C′(9, 21)
2. B. A′(0, 15), B′(−6, 18), C′(9, 24)
3. C. A′(0, 9), B′(−6, 12), C′(9, 18)
4. D. A′(0, 4), B′(−6, 7), C′(9, 13)

Answer 1

Hi Jennifer,

First I suggest that you plot the vertices of triangle ABC on a graph paper.

You have to dilate by a factor of 3 about the vertex A.

To do this

1. mark the center of dilation which is vertex A (0,4)

2.Find the distance of each vertex in your original (pre-image) from center of dilation

3.Move each vertex with respect to the center of dilation by Multiplying its distance by scale factor 3 and plot their new positions.

For Eg, Vertex B (-2,5) - To get to B from center of dilation (0,4) we have to move 2 units left and1 unit up. Now multiply this 3 times to get to new point B', so you would move 6 units left and 3 units up. So the coordinates of dilated point B' would be (-6,3).

I will let you do the other two vertices.

4.You will notice that the distance of vertex A from the center of dilation is 0, so even if you multiplied by 3 its distance would be 0, so it has to remain in its original position.

Hope this helps!