Find L on A (-2,0) and B(4,3) that is 2/3 of the distance from A to B. A. (0,1)

B (1,3/2)

C(2/3, 3/2)

D(2,2)

Greetings from Brasil...

First we will calculate the distance between points A and B....

The expression that allows you to calculate the distance between two points is:

d(A; B) = √[(4 - (- 2))² + (3 - 0)²]

d(A; B) = √[6² + 3²]

45 = 3² · 5

d(A; B) = 3√5

L is 2/3 of the distance from AB, so

L = 2/3 · d(A; B)

L = 2/3 · 3√5

For the coordinates of point L: let us use similarity of triangles...

BM/LN = AB/AL     - refer to attached

3/Yl = 3√5/2√5

Yl = 2

As we can see in the graph, when Y = 2, X is also 2, so

there are other ways to accomplish this task

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.