+0

Dilations and Similarity in the Coordinate Plane

0
718
3
+5259

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

Prove: Triangle ABC is similar to Triangle ADE

Mar 3, 2016

#1
+2496
+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:

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

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Mar 4, 2016

#1
+2496
+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
+105509
+20

Nice work solveit :)

Mar 5, 2016
#3
+105509
+15

Solveit,

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

AB = 2*sqrt(10)

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  :))

Mar 6, 2016