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Standard XII
Mathematics
Question
The area of a triangle with vertices
(
a
,
b
+
c
)
,
(
b
,
c
+
a
)
and
(
c
,
a
+
b
)
is
a
b
c
0
A
a
B
0
C
c
D
b
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Solution
Verified by Toppr
The area of
△
A
B
C
with vertices
A
≡
(
x
1
,
y
1
)
,
B
≡
(
x
2
,
y
2
)
and
C
≡
(
x
3
,
y
3
)
is given as,
A
(
△
A
B
C
)
=
∣
∣
∣
1
2
[
x
1
(
y
3
−
y
2
)
+
x
2
(
y
1
−
y
3
)
+
x
3
(
y
2
−
y
1
)
]
∣
∣
∣
In this problem,
A
(
△
A
B
C
)
=
∣
∣
∣
1
2
[
a
(
a
+
b
−
(
c
+
a
)
)
+
b
(
b
+
c
−
(
a
+
b
)
)
+
c
(
c
+
a
−
(
b
+
c
)
)
]
∣
∣
∣
∴
A
(
△
A
B
C
)
=
∣
∣
∣
1
2
[
a
b
−
a
c
+
b
c
−
b
a
+
c
a
−
c
b
]
∣
∣
∣
∴
A
(
△
A
B
C
)
=
0
Hence, the correct Option is B.
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15
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Q1
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,
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Q2
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Q3
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