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AP EAMCET Engineering 2012 Solved Paper
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© examsnet.com
Question : 83
Total: 160
The distance between the parallel lines given by
(
x
+
7
y
)
2
+
4
√
2
(
x
+
7
y
)
−
42
=
0
is
4
5
4
√
2
2
10
√
2
Validate
Solution:
Given equation is
(
x
+
7
y
)
2
+
4
√
2
(
x
+
7
y
)
−
42
=
0
Put
x
+
7
y
=
t
⇒
t
2
+
4
√
2
(
t
)
−
42
=
0
⇒
t
=
−
4
√
2
±
√
32
+
168
2
=
−
4
√
2
±
10
√
2
2
=
3
√
2
,
−
7
√
2
∴
x
+
7
y
=
3
√
2
and
x
+
7
y
=
−
7
√
2
⇒
x
+
7
y
−
3
√
2
=
0
and
x
+
7
y
+
7
√
2
=
0
∴ Distance between parallel lines
=
|
7
√
2
+
3
√
2
|
√
1
2
+
7
2
=
10
√
2
5
√
2
=
2
© examsnet.com
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