Vanessa draws one side of an equilateral triangle ABC on the coordinate plane at points A(-2, 1) and B (4, 1). What are the two possible coordinates of vertex C? Round to the nearest tenth.

two vertex are: (1 , 6.2) , (1 , - 4.2)

Step-by-step explanation:

A(-2, 1) and B (4, 1)    AB is a parallel segment with  x axis

The 3rd vertex P (x , y) will locate at the mid point of AB

x = (-2) + (4 - (-2)) /2 = 1       P (1 , y)

length AB = 4 - (-2) = 6

AP = AB (equilateral)

BP² = (4 - 1)² + (1 - y)² = 6²

1 - 2y + y² = 36 - 9 = 27

y² - 2y -26 = 0

y = ((-(-2)) ± √(-2)² - 4 x (-26)) / 2 = (2 ± √108) / 2 = 1 ± 5.2

y = 6.2 or y = -4.2

P (1 , 6.2) or P ( 1 , -4.2)

hi,

i'm 75% sure the answer is 1,1 to both lines.

have a great day

∠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.