Dilations and Similarity in the Coordinate Plane

what you need to find is distance from one point to another (sides of triangle):

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \text{ABC}_{triangle}\\ AB=\sqrt {(-6)^2 + (-2)^2}= 6.324555\\ AC=\sqrt {(4 - 2)^2 + (-2 - 4)^2}= 6.324555\\ BC=\sqrt {(4 - (-4))^2 + (-2 - 2)^2}=8.944272$

$ADE_{triangle}\\ AD=\sqrt {(-1 - 2)^2 + (3 - 4)^2}=3.162278\\ AE=\sqrt {(3 - 2)^2 + (1 - 4)^2}=3.162278\\ DE=\sqrt {(3 - (-1))^2 + (1 - 3)^2}=4.472136$

SO:

$\frac{AB}{AD}=\frac{AC}{AE}=\frac{BC}{DE}=1.999999683772268$

Nice work solveit :)

Solveit,

Your answer is very good but you would have been better off to stick to exact values.

AD =sqrt (10)

AB = 2*sqrt(10)

so     AB/AD = 2

I am sure that this would be true for the other ratios that you looked at as well.

So the triangles are similar AND the dimensions of the bigger one are exactly twice that of the smaller one  :))