What is the equation of the line that passes through the point (-6, 2) and has a
slope of -3/2?

y= -3/2x -7

Step-by-step explanation:

Since the question does not specify, we can answer in slope intercept form which is y=mx+b

use the slope and coord. points to find b

slope: -3/2, point: (-6, 2), -6 is your x coord, and 2 is your y coord.

2= -3/2(-6) + b

2= 18/2 + b Note: negative times a negative makes a positive.

2= 9+b subtract 9 from both sides to get b alone.

-9  -9

b=-7 now that we have b, the equation of the line would be

y= -3/2x -7

∠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.