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Warm Up Construct each of the following. 1. A perpendicular bisector. 2. An angle bisector. 3. Find the midpoint and slope of the segment (2, 8) and (–4, 6).

Warm Up Construct each of the following. 1. A perpendicular bisector. 2. An angle bisector

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Warm Up Construct each of the following. 1. A perpendicular bisector. 2. An angle bisector. 3. Find the midpoint and slope of the segment (2, 8) and (–4, 6). When a point is the same distance from two or more objects, the point is said to be equidistant from - PowerPoint PPT Presentation

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Page 1: Warm Up Construct each of the following. 1. A perpendicular bisector. 2.  An angle bisector

Warm UpConstruct each of the following.

1. A perpendicular bisector.

2. An angle bisector.

3. Find the midpoint and slope of the segment

(2, 8) and (–4, 6).

Page 2: Warm Up Construct each of the following. 1. A perpendicular bisector. 2.  An angle bisector

When a point is the same distance from two or moreobjects, the point is said to be equidistant fromthe objects. Triangle congruence theorems can beused to prove theorems about equidistant points.

Page 3: Warm Up Construct each of the following. 1. A perpendicular bisector. 2.  An angle bisector

A locus is a set of points that satisfies a given condition. The perpendicular bisector of a segment can be defined as the locus of points in a plane that are equidistant from the endpoints of the segment.

Example 1A: Find each measure.

MN

MN = LN

MN = 2.6

Bisector Thm.

Substitute 2.6 for LN.

Remember that the distance between a point and a line is the length of the perpendicular segment from the point to the line.

Page 4: Warm Up Construct each of the following. 1. A perpendicular bisector. 2.  An angle bisector

Example 1B:

TU

Find each measure.

So TU = 3(6.5) + 9 = 28.5.

TU = UV Bisector Thm.

3x + 9 = 7x – 17

9 = 4x – 17

26 = 4x

6.5 = x

Subtract 3x from both sides.

Add 17 to both sides.

Divide both sides by 4.

Substitute the given values.

Page 5: Warm Up Construct each of the following. 1. A perpendicular bisector. 2.  An angle bisector

Based on these theorems, an angle bisector can be defined as the locus of all points in the interior of the angle that are equidistant from the sides of the angle.

Page 6: Warm Up Construct each of the following. 1. A perpendicular bisector. 2.  An angle bisector

Example 2A: Find the measure.

BC

BC = DC

BC = 7.2

Bisector Thm.

Substitute 7.2 for DC.

Example 2B: Find mMKL.

mMKL = mJKM

3a + 20 = 2a + 26

a + 20 = 26

a = 6

Def. of bisector

Substitute the given values.

Subtract 2a from both sides.

Subtract 20 from both sides.

So mMKL = [2(6) + 26]° = 38°

Since, JM = LM, KJ JM and KL LM, KM bisects JKL by the Converse of the Angle Bisector Theorem.

Page 7: Warm Up Construct each of the following. 1. A perpendicular bisector. 2.  An angle bisector

Example 4: Write an equation in point-slope form for the perpendicular bisector of the segment with endpoints C(6, –5) and D(10, 1).

The perpendicular bisector of is perpendicular to at its midpoint.

Step 1 Find the midpoint of . Midpoint formula.

mdpt. of =

Step 2 Find the slope of the perpendicular bisector.

Slope formula.

Page 8: Warm Up Construct each of the following. 1. A perpendicular bisector. 2.  An angle bisector

Since the slopes of perpendicular lines are opposite reciprocals,

the slope of the perpendicular bisector is

Step 3 Use point-slope form to write an equation.

The perpendicular bisector of has slope and

passes through (8, –2).

y – y1 = m(x – x1) Point-slope form

Substitute –2 for y1,

for m, and 8 for

x1.