4The distance between any two points is the absolute value of the difference of the coordinates. If the coordinates of points A and B are a and b, then the distance between A and B is |a – b| or |b – a|. The distance between A and B is also called the length of AB, or AB.AB = |a – b| or |b - a|AaBb
5Congruent segments are segments that have the same length Congruent segments are segments that have the same length. In the diagram, PQ = RS, so you can write PQ RS. This is read as “segment PQ is congruent to segment RS.” Tick marks are used in a figure to show congruent segments.
6In order for you to say that a point B is between two points A and C, all three points must lie on the same line, and AB + BC = AC.
7Example 1: Using the Segment Addition Postulate M is between N and O. Find NO.NM + MO = NOSeg. Add. Postulate17 + (3x – 5) = 5x + 2Substitute the given values3x + 12 = 5x + 2Simplify.– – 2Subtract 2 from both sides.3x + 10 = 5xSimplify.–3x –3xSubtract 3x from both sides.10 = 2xDivide both sides by 2.25 = x
8Example 1 ContinuedM is between N and O. Find NO.NO = 5x + 2= 5(5) + 2Substitute 5 for x.= 27Simplify.
9Check It Out! Example 2E is between D and F. Find DF.DE + EF = DFSeg. Add. Postulate(3x – 1) + 13 = 6xSubstitute the given values3x + 12 = 6x– 3x – 3xSubtract 3x from both sides.12 = 3xSimplify.x3=Divide both sides by 3.4 = x
10Check It Out! Example 2 Continued E is between D and F. Find DF.DF = 6x= 6(4)Substitute 4 for x.= 24Simplify.
11The midpoint M of AB is the point that bisects, or divides, the segment into two congruent segments. If M is the midpoint of AB, then AM = MB.So if AB = 6, then AM = 3 and MB = 3.
12Example 3: Recreation Application The map shows the route for a race. You are at X, 6000 ft from the first checkpoint C. The second checkpoint D is located at the midpoint between C and the end of the race Y. The total race is 3 miles. How far apart are the 2 checkpoints?XY = 3(5280 ft)Convert race distance to feet.= 15,840 ft
13Example 3 ContinuedXC + CY = XYSeg. Add. Post.Substitute 6000 for XC and 15,840 for XY.CY = 15,840– – 6000Subtract 6000 from both sides.CY = 9840Simplify.D is the mdpt. of CY, so CD = CY.= 4920 ftThe checkpoints are 4920 ft apart.
14Example 4: Using Midpoints to Find Lengths D is the midpoint of EF, ED = 4x + 6, and DF = 7x – 9. Find ED, DF, and EF.ED4x + 67x – 9FStep 1 Solve for x.ED = DFD is the mdpt. of EF.4x + 6 = 7x – 9Substitute 4x + 6 for ED and 7x – 9 for DF.–4x –4xSubtract 4x from both sides.6 = 3x – 9Simplify.Add 9 to both sides.15 = 3xSimplify.
15Example 4 ContinuedD is the midpoint of EF, ED = 4x + 6, and DF = 7x – 9. Find ED, DF, and EF.ED4x + 67x – 9FStep 2 Find ED, DF, and EF.ED = 4x + 6DF = 7x – 9EF = ED + DF= 4(5) + 6= 7(5) – 9== 26= 26= 52
16Check It Out! Example 5S is the midpoint of RT, RS = –2x, andST = –3x – 2. Find RS, ST, and RT.RST–2x–3x – 2Step 1 Solve for x.RS = STS is the mdpt. of RT.–2x = –3x – 2Substitute –2x for RS and –3x – 2 for ST.+3x +3xAdd 3x to both sides.x = –2Simplify.
17Check It Out! Example 5 Continued S is the midpoint of RT, RS = –2x, andST = –3x – 2. Find RS, ST, and RT.RST–2x–3x – 2Step 2 Find RS, ST, and RT.RS = –2xST = –3x – 2RT = RS + ST= –2(–2)= –3(–2) – 2= 4 + 4= 4= 4= 8
18The set of all points between the sides of the angle is the interior of an angle. The exterior of an angle is the set of all points outside the angle.Angle NameR, SRT, TRS, or 1You cannot name an angle just by its vertex if the point is the vertex of more than one angle. In this case, you must use all three points to name the angle, and the middle point is always the vertex.
19Check It Out! Example 6Write the different ways you can name the angles in the diagram.RTQ, T, STR, 1, 2
20The measure of an angle is usually given in degrees The measure of an angle is usually given in degrees. Since there are 360° in a circle, one degree is of a circle. When you use a protractor to measure angles, you are applying the following postulate.
21You can use the Protractor Postulate to help you classify angles by their measure. The measure of an angle is the absolute value of the difference of the real numbers that the rays correspond with on a protractor.If OC corresponds with c and OD corresponds with d,mDOC = |d – c| or |c – d|.
23Check It Out! Example 7Use the diagram to find the measure of each angle. Then classify each as acute, right, or obtuse.a. BOAb. DOBc. EOCmBOA = 40°BOA is acute.mDOB = 125°DOB is obtuse.mEOC = 105°EOC is obtuse.
24Congruent angles are angles that have the same measure Congruent angles are angles that have the same measure. In the diagram, mABC = mDEF, so you can write ABC DEF. This is read as “angle ABC is congruent to angle DEF.” Arc marks are used to show that the two angles are congruent.The Angle Addition Postulate is very similar to the Segment Addition Postulate that you learned in the previous lesson.
26Check It Out! Example 8mXWZ = 121° and mXWY = 59°. Find mYWZ.mYWZ = mXWZ – mXWY Add. Post.mYWZ = 121 – 59Substitute the given values.mYWZ = 62Subtract.
27An angle bisector is a ray that divides an angle into two congruent angles. JK bisects LJM; thus LJK KJM.
28Example 9: Finding the Measure of an Angle KM bisects JKL, mJKM = (4x + 6)°, and mMKL = (7x – 12)°. Find mJKM.
29Example 9 ContinuedStep 1 Find x.mJKM = mMKLDef. of bisector(4x + 6)° = (7x – 12)°Substitute the given values.Add 12 to both sides.4x = 7xSimplify.–4x –4xSubtract 4x from both sides.18 = 3xDivide both sides by 3.6 = xSimplify.
31Check It Out! Example 10Find the measure of each angle.QS bisects PQR, mPQS = (5y – 1)°, andmPQR = (8y + 12)°. Find mPQS.Step 1 Find y.Def. of bisectorSubstitute the given values.5y – 1 = 4y + 6Simplify.y – 1 = 6Subtract 4y from both sides.y = 7Add 1 to both sides.
32Check It Out! Example 10 Continued Step 2 Find mPQS.mPQS = 5y – 1= 5(7) – 1Substitute 7 for y.= 34Simplify.
33Check It Out! Example 11Find the measure of each angle.JK bisects LJM, mLJK = (-10x + 3)°, andmKJM = (–x + 21)°. Find mLJM.Step 1 Find x.LJK = KJMDef. of bisector(–10x + 3)° = (–x + 21)°Substitute the given values.+x xAdd x to both sides.Simplify.–9x + 3 = 21–3 –3Subtract 3 from both sides.–9x = 18Divide both sides by –9.x = –2Simplify.
34Check It Out! Example 11 Continued Step 2 Find mLJM.mLJM = mLJK + mKJM= (–10x + 3)° + (–x + 21)°= –10(–2) + 3 – (–2) + 21Substitute –2 for x.=Simplify.= 46°