The given line segment has a midpoint at (3, 1). what is the equation, in slope-intercept form, of the perpendicular bisector of the given line segment? y = x y = x – 2 y = 3x y = 3x − 8

The equation, in slope-intercept form, of the perpendicular bisector is: y = 1/3x.

The slope-intercept form is given as, y = mx + b.

b is y-intercept, and m is the slope.

Slope = change in y/change in x.

The slopes of two lines that are perpendicular to each other are negative reciprocal of each other.

Thus, let's find the slope of the blue line:

Slope = (-2 - 4)/(4 - 2) = -3

Therefore, the slope of the line perpendicular to it would be, 1/3.

Equation of the perpendicular bisector that passes through (3, 1):

Since slope (m) is 1/3, find b by substituting m = 1/3 and (x, y) = (3, 1) into y = mx + b.

Thus:

1 = 1/3(3) + b

1 = 1 + b

1 - 1 = b

b = 0.

Substitute b = 0, and m = 1/3 into y = mx + b:

y = 1/3x + 0

y = 1/3x

Therefore, the equation, in slope-intercept form, of the perpendicular bisector is: y = 1/3x.

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∠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.

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step-by-step explanation: