Suppose you have AB on a coordinate plane located at A(-3,-4) and B(5,-4)Under a dilation centered at (9,0), AB becomes A'B' with coordinates A' (6,-1)and B'(8,-1). What is the scale factor for this dilation?What’s the answer

Accepted Solution

Answer:The scale factor is 1/4Step-by-step explanation:There are two ways to get this answer:1. Graphically: After representing the point in the graph, you can count the distance between each point on the x and y axis, and the dilation point D). The distance in the y-axis between A'B' and D is 1, while AB and D is 4.The distance in the x-axis between A and D is 12, while A' and D is 3.We can see that in both axis the distance is reduced by 1/4.[tex](y_{axis} =\frac{1}{4}); (x_{axis} =\frac{3}{12}=\frac{1}{4})[/tex]2. Mathematically: To do this is necessary to find the distance between for example point A and A' with D, and then find the proportion [tex]\frac{A'}{A}[/tex][tex]d_{AD}=\sqrt{(X_{D}-X_{A})^{2} +(Y_{D}-Y_{A})^{2} }[/tex][tex]d_{AD}=\sqrt{(9-(-3))^{2} +(0-(-4))^{2} } =\sqrt{(12)^{2} +(4)^{2} }[/tex][tex]d_{AD}=\sqrt160} =4\sqrt{10}[/tex]Following the same procedure:[tex]d_{A'D}=\sqrt{(9-6)^{2} +(0-(-1))^{2} }=\sqrt{(3)^{2} +(1)^{2} }=\sqrt{10}[/tex]Therefore, the proportion is [tex]\frac{d_{A'D}}{d_{AD}}=\frac{\sqrt{10}}{4\sqrt{10}} =\frac{1}{4}[/tex]