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Unformatted text preview: 0.0 points Show that the triangle ∆ABC with vertices
at
A = (a1 , a2 ), B = (b1 , b2 ), C = (c1 , c2 )
has a1
1 b1
det
Area =
2
c1 a2
b2
c2 1
1 .
1 (Hint: translate ∆ABC so that A becomes
the origin.)
Explanation:
After translating ∆ABC so that A becomes
the origin, we obtain a new triangle ∆OB ′ C ′
of equal area with vertices at the origin and
B ′ = (b1 − a1 , b2 − a2 ), C ′ = (c1 − a1 , c2 − a2 ).
Now
Area(∆OB ′ C ′ )
= 1
b − a1
det 1
c1 − a1
2 b2 − a 2
c2 − a2 . On the other hand, by properties of determinants, a1 a2 1
det b1 b2 1 c1 c2 1 a1
a2
1
= det b1 − a1 b2 − a − 2 0 c1
c2
1 a1
a2
1
= det b1 − a1 b2 − a − 2 0 c1 − a1
c2 − a2
0
= det b1 − a 1
c1 − a1 b2 − a 2
c2 − a2 Consequently, ∆ABC has a1 a2
1 b1 b2
det
Area =
2
c1 c2 1
1
1 . . 5...
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This document was uploaded on 03/16/2014 for the course M 340L at University of Texas at Austin.
 Spring '08
 PAVLOVIC
 Matrices

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