Area of triangle whose vertices (4,2),(1,3)and(2,3)is
A)0.5
B)1.5
C)3.5
D)4.5
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Answer:
Let the vertices of triangle are
A(1,2,3),B(2,5,−1) and C(−1,1,2)
Then,
AB
=
OB
−
OA
=
i
^
+3
j
^
−4
k
^
AC
=
OC
−
OA
=−2
i
^
−
j
^
−
k
^
Then,
AB
×
AC
=
∣
∣
∣
∣
∣
∣
∣
∣
i
^
1
−2
j
^
3
−1
k
^
−4
−1
∣
∣
∣
∣
∣
∣
∣
∣
=−7
i
^
+9
j
^
+5
k
^
∣
∣
∣
∣
AB
×
AC
∣
∣
∣
∣
=
(−7)
2
+(9)
2
+(5)
2
=
49+8+25
=
155
Area of triangle ABC=
2
1
∣
AB
×
AC
∣
=
2
1
×
155
=
2
155
sq. units
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