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JEE Main 2013 Question Paper
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© examsnet.com
Question : 39
Total: 90
The x-coordinate of the in centre of the triangle that has the coordinates of mid points of its sides as (0, 1) (1, 1) and (1, 0) is
2
−
√
2
1
+
√
2
1
−
√
2
2
+
√
2
Validate
Solution:
<-coordinate
=
a
x
1
+
b
x
2
+
c
x
3
a
+
b
+
c
=
2
×
2
+
2
√
2
×
0
+
2
×
0
2
+
2
+
2
√
2
=
4
4
+
2
√
2
=
2
2
+
2
√
=
2
−
√
2
Alternate Solution:
x-coordinate = r = ( s - a ) tan A/2
=
(
4
+
2
√
2
2
−
2
√
2
)
t
a
n
π
4
=
2
−
√
2
© examsnet.com
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