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# Dilations and Similarity in the Coordinate Plane

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307
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Given: A(2, 4), B(-4, 2), C(4, -2), D(-1, 3), E(3, 1)

Prove: Triangle ABC is similar to Triangle ADE

rarinstraw1195  Mar 3, 2016

#1
+2493
+25

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:

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

Solveit  Mar 4, 2016
Sort:

#1
+2493
+25

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:

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

Solveit  Mar 4, 2016
#2
+91469
+20

Nice work solveit :)

Melody  Mar 5, 2016
#3
+91469
+15

Solveit,

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

AB = 2*sqrt(10)