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MHT CET 2020 Math Solved Paper
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© examsnet.com
Question : 29
Total: 50
The locus of a point that is equidistant from the lines
x
+
y
−
2
√
2
=
0
and
x
+
y
−
√
2
=
0
is
x
+
y
−
5
√
2
=
0
x
+
y
−
3
√
2
=
0
2
x
+
2
y
−
3
√
2
=
0
2
x
+
2
y
−
5
√
2
=
0
Validate
Solution:
For any point
P
(
x
,
y
)
that is equidistant from the given line, we have
x
+
y
−
√
2
=
−
(
x
+
y
−
2
√
2
)
or
2
x
+
2
y
−
3
√
2
=
0
.
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