Unformatted text preview: 30. Use the Distance Formula to find DE and AB. 26. In order to show QT RS, we must show that
PQ
QR
PQ
QR
PT
TS PT
TS .
¬ 34.88 or 436
18.32
229
33.25 11.45
¬
11.45
21.8
¬ 11.45 or 436
229
PQ
PT
436
Thus, QR
TS
229 . Since the sides have y A 4
x 8 B DE ¬3
2 , 26 3 2 4 1
2 2 81
4 10 31. y C 1
2 AB. A (2, 12) (5, 17). 20 C ¬ [ 4 ( 1)]2 ( 3 6)2
¬ 9 81
¬3 10
So, 3 10 1 (3 10) and thus DE
2
2 slope of TS
Because the slopes of WM and TS are equal,
WM TS.
WM is a midsegment of RST if W is the
midpoint of RT and M is the midpoint of RS.
11 8 4 AB 12 14
5 3 or 1
20 26
17 11 or 1 2 3
2 ¬
¬9
4 28. If the slopes of WM and TS are equal, WM TS. 1 8 8 27. DE is a midsegment of ABC and DE BC, so by
the Triangle Midsegment Theorem, DE 1 BC.
2
Then BC 2DE.
DE ¬ (4 1)2 (3 1)2
¬9 4
¬ 13
BC ¬2DE
¬2 13
¬ 4 13 or 52 The midpoint of RT is 4 4 proportional lengths, QT RS. slope of WM 8 (8, 14). 2 These are not the coordinates of W.
17 8 1 The midpoint of RS is , 2 2 D These are not the coordinates of M.
WM is not a midsegment because W and M are
not midpoints of their respective sides. y 8
4
x 8 4 4 B 8 4 C 8 D 1 76
2, E 47
2, ( 5)
2
3 x Graph AB. We can find segments of AB with
lengths in a ratio of 2 to 1 by considering a second
line and parallel lines that intersect this line and
AB.
Graph C(0, 12) and D(0, 0) and lines CA and DB.
CA and DB are horizontal lines and are parallel
lines intersecting transversals CD (the yaxis) and
AB. We can find P by finding a third parallel line
intersecting CD and AB so that this line
separates CD into two parts with a ratio of 2 to 1.
CD is 12 units, so if a horizontal line intersects
CD at (0, 4) or (0, 8) then this line separates CD
into two parts with a ratio of 2 to 1. These
horizontal lines intersect AB at (4, 4) and (3, 8).
These points cut off AB into parts with a ratio of
2 to 1, so P could have coordinates (4, 4) or (3, 8). 29. DE is a midsegment of ABC. Use the Midpoint
Formula to find the coordinates of D and E. A B (5, 0) D 3, 1
2 ( 5) E 3,
2 2 4 Find the slopes of DE and AB.
slope of DE
slope of AB 4
3
2 or 3 3
3 4 1
2 6
( 1) or 3 Both DE and AB have slope 3, so DE is parallel
to AB. 171 Chapter 6 ...
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This note was uploaded on 12/19/2011 for the course MAC 1140 taught by Professor Dr.zhan during the Fall '10 term at UNF.
 Fall '10
 Dr.Zhan
 Calculus, PreCalculus, Distance Formula

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