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Section:
Mathematics
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© examsnet.com
Question : 10 of 52
Marks:
+1
,
-0
Let
X
=
{
(
x
,
y
)
∈
Z
×
Z
:
x
2
8
+
y
2
20
<
1.
X = \{ (x, y) \in \mathbb{Z} \times \mathbb{Z} : \frac{x^2}{8} + \frac{y^2}{20} < 1.
X
=
{(
x
,
y
)
∈
Z
×
Z
:
8
x
2
+
20
y
2
<
1.
and
y
2
<
5
x
}
y^2 < 5 x \}
y
2
<
5
x
}
. Three distinct point
P
,
Q
P, Q
P
,
Q
and
R
R
R
are randomly chosen from
X
X
X
. Then the probability that
P
,
Q
P, Q
P
,
Q
and
R
R
R
form a triangle whose area is a positive integer, is
[JEE Adv 2023 P1]
71
220
\frac{71}{220}
220
71
73
220
\frac{73}{220}
220
73
79
220
\frac{79}{220}
220
79
83
220
\frac{83}{220}
220
83
Validate
Solution:
The given region are as
The points inside region are
{
(
2
,
1
)
,
(
2
,
−
1
)
,
(
2
,
2
)
,
(
2
,
−
2
)
,
(
2
,
3
)
,
(
2
,
−
3
)
,
(
2
,
0
)
,
(
1
,
1
)
,
(
1
,
−
1
)
,
(
1
,
2
)
,
(
1
,
−
2
)
\{(2,1),(2,-1),(2,2),(2,-2),(2,3),(2,-3),(2,0),(1,1),(1,-1),(1,2),(1,-2)
{(
2
,
1
)
,
(
2
,
−
1
)
,
(
2
,
2
)
,
(
2
,
−
2
)
,
(
2
,
3
)
,
(
2
,
−
3
)
,
(
2
,
0
)
,
(
1
,
1
)
,
(
1
,
−
1
)
,
(
1
,
2
)
,
(
1
,
−
2
)
,
(
1
,
0
)
}
(1,0)\}
(
1
,
0
)}
.
Total number of ways to select three points
=
12
C
3
=
220
= {}^{12}C_3 = 220
=
12
C
3
=
220
Required number of triangle
=
4
×
7
C
1
+
9
×
5
C
1
=
73
= 4 \times {}^{7}C_1 + 9 \times {}^{5}C_1 = 73
=
4
×
7
C
1
+
9
×
5
C
1
=
73
Points are taken such a way that distance between two points are multiple of 2 .
© examsnet.com
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