Objectives Use length and midpoint of a segment.

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Presentation transcript:

Objectives Use length and midpoint of a segment. Construct midpoints and congruent segments.

The 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| A a B b

Example 1: Finding the Length of a Segment Find each length. A. BC B. AC BC = |1 – 3| AC = |–2 – 3| = |– 2| = |– 5| = 2 = 5

Check It Out! Example 1 Find each length. a. XY b. XZ

Congruent 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.

In 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.

Example 3A: Using the Segment Addition Postulate G is between F and H, FG = 6, and FH = 11. Find GH. FH = FG + GH Seg. Add. Postulate 11 = 6 + GH Substitute 6 for FG and 11 for FH. – 6 –6 Subtract 6 from both sides. 5 = GH Simplify.

Example 3B: Using the Segment Addition Postulate M is between N and O. Find NO. NM + MO = NO Seg. Add. Postulate 17 + (3x – 5) = 5x + 2 Substitute the given values 3x + 12 = 5x + 2 Simplify. – 2 – 2 Subtract 2 from both sides. 3x + 10 = 5x Simplify. –3x –3x Subtract 3x from both sides. 10 = 2x Divide both sides by 2. 2 5 = x

Example 3B Continued M is between N and O. Find NO. NO = 5x + 2 = 5(5) + 2 Substitute 5 for x. = 27 Simplify.

Check It Out! Example 3a Y is between X and Z, XZ = 3, and XY = . Find YZ. XZ = XY + YZ Seg. Add. Postulate Substitute the given values. Subtract from both sides.

Check It Out! Example 3b E is between D and F. Find DF. DE + EF = DF Seg. Add. Postulate (3x – 1) + 13 = 6x Substitute the given values 3x + 12 = 6x – 3x – 3x Subtract 3x from both sides. 12 = 3x Simplify. 12 3x 3 = Divide both sides by 3. 4 = x

Check It Out! Example 3b Continued E is between D and F. Find DF. DF = 6x = 6(4) Substitute 4 for x. = 24 Simplify.

The 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.

Example 5: Using Midpoints to Find Lengths D is the midpoint of EF, ED = 4x + 6, and DF = 7x – 9. Find ED, DF, and EF. E D 4x + 6 7x – 9 F Step 1 Solve for x. ED = DF D is the mdpt. of EF. 4x + 6 = 7x – 9 Substitute 4x + 6 for ED and 7x – 9 for DF. –4x –4x Subtract 4x from both sides. 6 = 3x – 9 Simplify. +9 + 9 Add 9 to both sides. 15 = 3x Simplify.

Example 5 Continued D is the midpoint of EF, ED = 4x + 6, and DF = 7x – 9. Find ED, DF, and EF. E D 4x + 6 7x – 9 F 3 3 15 3x = Divide both sides by 3. x = 5 Simplify.

Example 5 Continued D is the midpoint of EF, ED = 4x + 6, and DF = 7x – 9. Find ED, DF, and EF. E D 4x + 6 7x – 9 F Step 2 Find ED, DF, and EF. ED = 4x + 6 DF = 7x – 9 EF = ED + DF = 4(5) + 6 = 7(5) – 9 = 26 + 26 = 26 = 26 = 52

Check It Out! Example 5 S is the midpoint of RT, RS = –2x, and ST = –3x – 2. Find RS, ST, and RT. R S T –2x –3x – 2 Step 1 Solve for x. RS = ST S is the mdpt. of RT. –2x = –3x – 2 Substitute –2x for RS and –3x – 2 for ST. +3x +3x Add 3x to both sides. x = –2 Simplify.

Check It Out! Example 5 Continued S is the midpoint of RT, RS = –2x, and ST = –3x – 2. Find RS, ST, and RT. R S T –2x –3x – 2 Step 2 Find RS, ST, and RT. RS = –2x ST = –3x – 2 RT = RS + ST = –2(–2) = –3(–2) – 2 = 4 + 4 = 4 = 4 = 8

Practice Quiz: Part I 1. M is between N and O. MO = 15, and MN = 7.6. Find NO. 22.6 2. S is the midpoint of TV, TS = 4x – 7, and SV = 5x – 15. Find TS, SV, and TV. 25, 25, 50

Practice Quiz: Part II 3. LH bisects GK at M. GM = 2x + 6, and GK = 24. Find x. 3 4. Tell whether the statement below is sometimes, always, or never true. Support your answer with a sketch. If M is the midpoint of KL, then M, K, and L are collinear. Always K L M