What is an equation in slope-intercept form of the line that passes through (6, −7) and is perpendicular to the line shown below?

A.

Step-by-step explanation:

Equation of a Line

The equation of the line in slope-intercept form is:

y=mx+b

Where:

m = slope

b  = y-intercept.

The point-slope form of the equation of a line is:

y - k = m ( x - h )

Where:

(h,k) is a point through which the line passes.

The line we are looking for has a slope defined for the fact that is perpendicular to the line shown in the graph.

Two perpendicular lines with slopes m1 and m2 satisfy the equation:

We'll find the slope m1 of the given line and then solve the above equation for m2:

The line of the graph passes through two clear points (-3,-3) and (0,3). Let's calculate the slope.

Suppose we know the line passes through points A(x1,y1) and B(x2,y2). The slope can be calculated with the equation:

Now we calculate the second slope:

We use the point-slope form, given the point (6,-7):

Operating the parentheses:

Simplifying:

A.

∠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.

2: 10

step-by-step explanation: