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Byju's Answer
Standard XII
Mathematics
Area of Triangle Using Coordinates
The sides of ...
Question
The sides of a triangle are
x
=
2
,
y
+
1
=
0
and
x
+
2
y
=
4
. Its circumcentre is-
A
(
4
,
0
)
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B
(
2
,
−
1
)
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C
(
0
,
4
)
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D
(
2
,
3
)
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Solution
The correct option is
B
(
4
,
0
)
sides are
x
=
2
,
y
=
−
1.
x
+
2
y
=
4
Solving these we get get vertices of the
Triangle are
A
(
2
,
−
1
)
,
B
(
2
,
1
)
,
C
(
6
,
−
1
)
Let circumcentre is
d
(
x
,
y
)
circumcentre is equidistant from all three vertices so
C
′
A
=
C
′
B
(
x
−
2
)
2
+
(
y
+
1
)
2
=
(
x
−
2
)
2
+
(
y
−
1
)
2
y
=
0
C
′
B
=
C
′
C
(
x
−
2
)
2
+
(
y
−
1
)
2
=
(
x
−
6
)
2
+
(
y
+
1
)
2
P
u
t
y
=
0
(
x
−
2
)
2
+
1
=
(
x
−
6
)
2
+
1
x
−
2
=
±
(
x
−
6
)
x
−
2
=
x
−
6
Invalid
x
−
2
=
−
(
x
−
6
)
2
x
=
8
x
=
4
So, circumcentre is
C
′
(
4
,
0
)
Suggest Corrections
0
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