The smaller triangle is the image of the larger triangle after a dilation. The center of the dilation is (4,2).https://static.k12.com/nextgen_media/assets/8080739-NG_GMT_SemA_06_UT_02.png, (DOUBLE CLICK IT AND THEN CLICK WWW.ETC, THEN AN IMAGE WILL POP UP.

Given that the smaller triangle is the image of the larger triangle after a dilation and that the center of the dilation is (4,2).

Let the scale factor used to create the dilation be k, then the distance of any point of the smaller triangle from the center of dilation is k times the distance of the corresponding point of the bigger triangle from the centre of dilation. Also, the size of any part of the smaller triangle is k times the size of the corresponding part of the bigger triangle.

Consider the hypothenuse of the two triangles, the length of the hypothenuse of the bigger triangle is given by

[tex]l_b= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ = \sqrt{(4-(-8))^2+(-7-2)^2} = \sqrt{(4+8)^2+(-9)^2} \\ \\ = \sqrt{12^2+81} = \sqrt{144+81} = \sqrt{225} =15[/tex]

while the length of the hypothenuse of the smaller triangle is given by

[tex]l_s= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ = \sqrt{(8-4)^2+(2-5)^2} = \sqrt{4^2+(-3)^2} \\ \\ = \sqrt{16+9} = \sqrt{25} =5[/tex]

Now,

[tex]l_s=kl_b \\ \\ \Rightarrow5=15k \\ \\ \Rightarrow k= \frac{5}{15} = \frac{1}{3} [/tex]

Therefore, the scale factor used to create the dilation is 1/3.