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Byju's Answer
Standard X
Mathematics
Area of a Triangle Given Its Vertices
Find the ince...
Question
Find the incentre of the triangle whose vertices are
(
2
,
3
)
,
(
−
2
,
−
5
)
,
(
−
4
,
6
)
.
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Solution
Let
A
=
(
2
,
3
)
B
=
(
−
2
,
−
5
)
C
=
(
−
4
,
6
)
Now by using straight line formula for two points
√
(
x
1
−
x
2
)
2
+
(
y
1
−
y
2
)
2
We get,
B
C
=
√
(
x
1
−
x
2
)
2
+
(
y
1
−
y
2
)
2
=
√
(
−
2
+
4
)
2
+
(
−
5
−
6
)
2
=
√
2
2
+
(
−
11
)
2
=
√
4
+
121
=
√
125
=
5
√
5
Now , again we get
A
C
=
√
(
2
+
4
)
2
+
(
6
−
3
)
2
=
√
6
2
+
3
2
=
√
36
+
9
=
√
45
=
3
√
5
similarly,
A
B
=
√
(
2
+
2
)
2
+
(
3
+
5
)
2
=
√
4
2
+
8
2
=
√
16
+
64
=
√
80
=
4
√
5
Now for the incentre of the triangle we get
I
=
(
a
x
1
+
b
x
2
+
c
x
3
a
+
b
+
c
,
a
y
1
+
b
y
2
+
c
y
3
a
+
b
+
c
)
=
(
5
√
5
×
2
+
3
√
5
×
(
−
2
)
+
4
√
5
×
(
−
4
)
5
√
5
+
3
√
5
+
4
√
5
,
5
√
5
×
3
+
3
√
5
×
(
−
5
)
+
4
√
5
×
6
5
√
5
+
3
√
5
+
4
√
5
)
=
(
10
√
5
−
6
√
5
−
6
√
5
12
√
5
,
15
√
5
−
15
√
5
+
24
√
5
12
√
5
)
=
(
−
12
√
5
12
√
5
,
24
√
5
12
√
5
)
=
(
−
1
,
2
)
Hence the required points of incentre is
=
(
−
1
,
2
)
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0
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