...

Document 1804691

by user

on
Category: Documents
59

views

Report

Comments

Transcript

Document 1804691
Challenge Problems for 3.1-3.3
Remember points of concurrency from geometry? lf you haven't completely blocked that part of your mathematical education
from your memory, you may recall that three or more lines are concurrent if they all intersect at the same point. The point that
they all have in common is called the point of concurrency. What we discovered in geometry is that there are a number of
points of concurrency asSociated with triangles: Centroid {where the medians intersect}, Circumcenter (where the
perpendicular bisectors intersect), lncenter (where the angle bisectors intersect), and Orthocenter (where the altitudes
intersect).
Points of Concurrency for Triangles
lncenter
Circumcenter
Centroid
-
lntersection of the medians
Median: segment that
connects a vertex to the
mldpoint of the opposite side
Point
C
*
-
lntersection ofthe
perpendicular bisectors
Perp. Bisector: line that
intersects a side at its midpoint
and is perpendicular to it
Point
-
Orthocenter
-
lntersection ofthe angle
bisectors
Anqle Bisector: segment that
bisects an angle
*
Point
lntersection of the altitudes
Altitude: a segment that starts
at a vertex and is
perpendicular to a line
containing the other side of
the triangle
I
C2
Point O
Find the indicated point of concurrency.
1.
ThenfindthecoordinatesoftheCentroid.
Findtheequationsofthethreemediansforthetrianglebelow.
frr
As: yrl
,!-s=*$G+ 4-\
tl
"'-:
-- -3^-4
5.
Iv
i
II
LD: y*{:
i.4,./\i.*or
,
|
': )
<
1
x . 'J-l
.
JT
I
-.;i*'
,lJl
i&-&:,i' {*l ,*".*
4
.t1
= --?
"J
l
; ": :r -?-=
:ti\"".4[
i" *'_-"-_*"J
&."-
2.
Find the equations of the three perpendicular bisectors of the sides of the triangle below. Then find the coordinates for the
Circumcenter
ABrgg
'
1:
--'--A
Cr): ,,4--'n-Y
€t-
Lr Y-il -;; r+,,
*11'
i
,--*,
lu i--:-J
tirriniry r{$'*' '
ft:)i
3.
Find the equations of the three altitudes for the triangle
1 t* ,J
4E)
find the coordinates for the Orthocenter'
Tl:*rll""
I 4
l
-_-*-/_\
"x +b\
CD', :Y:
\t*-*-'
ilttr"-t rdco:
f
I /
-
t'\
T
fF-\4-7=-i(x-v;
-_lx+1
"
4/\
+--":'
zb
\
V-.^:a
Fly UP