Write an equation in slope-intercept form of the line that passes through the given point and is parallel to the graph of the given equation.

(-3, 12); y = -3x + 5

y = -3x + 3 or y = 3 - 3x

Step-by-step explanation:

First, we need to determine our slope. Thankfully for us, the slope of a line parallel to another is the same.

So our slope is -3.

Now, we need to find our y-intercept. To do that, we will use y = mx + b. y is the y-coordinate, m is the slope, x is the x-coordinate, and b is the y-intercept.

Plug in the numbers given.

12 = -3(-3) + b

Simplify.

12 = 9 + b

Subtract 9 from both sides.

12 - 9 = 9 - 9 + b

3 = b

Our y-intercept is 3.

Now, we just put everything together.

y = -3x + 3

The line parallel to y = -3x + 5 that passes through the point (-3, 12) is

y = -3x + 3 or y = 3 - 3x (both are the same just written differently)

∠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.