Triangle DEF has vertices D(−4, 1), E(2, 3), and F(2, 1) and is dilated by a factor of 3 using the point (1, 1) as the point of dilation. The dilated triangle is named ΔD'E'F'. What are the coordinates of the vertices of the resulting triangle?begging someone for help please :(

Answer: [tex]D'(-14,1);\ E'(4,7);\ F'(4,1)[/tex] Step-by-step explanation: Since the center of dilation is not at the origin, we can use the following formula in order to find the coordinates of the vertices of the triangle D'E'F': [tex]D_{O,k}(x,y)=(k(x-a)+a, k(y-b)+b)[/tex] Where "O" is the center of dilation at (a,b) and "k" is the scale factor. In this case you can identify that: [tex](a,b)=(1,1)\\k=3[/tex] Therefore, susbtituting values into the formula shown above, you get that the coordinates ot the resulting triangle D'E'F, are the following: Vertex D' → [tex](3(-4-1)+1,\ 3(1-1)+1)=(-14,1)[/tex] Vertex E' → [tex](3(2-1)+1,\ 3(3-1)+1)=(4,7)[/tex] Vertex F' → [tex](3(2-1)+1,\ 3(1-1)+1)=(4,1)[/tex]