Which equation represents a line that passes through (2, –left-parenthesis 2, negative StartFraction one-half EndFraction right-parenthesis.) and has a slope of 3?

y – 2 = 3(x + y minus 3 equals 2 left-parenthesis x plus StartFraction one-half EndFraction right-parenthesis.)
y – 3 = 2(x + y minus 3 equals 2 left-parenthesis x plus StartFraction one-half EndFraction right-parenthesis.)
y + y plus StartFraction one-half EndFraction equals 3 left-parenthesis x minus 2 right-parenthesis. = 3(x – 2)
y + y plus StartFraction one-half EndFraction equals 2 left-parenthesis x minus 3 right-parenthesis. = 2(x – 3)

For this case we have that by definition, the equation of a line in the point-slope form is given by:

Where:

m: It is the slope of the line

It is a point through which the line passes

According to the data of the statement we have to:

We replace:

∠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.