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TG EAMCET 4 May 2025 Shift 2 Paper
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© examsiri.com
Question :
41 of 160
Marks:
+1
,
-0
A
(
2
,
0
)
,
B
(
0
,
2
)
,
C
(
−
2
,
0
)
are three points. Let
a
,
b
,
c
be the perpendicular distances from a variable point P on to the lines
AB
,
BC
and CA respectively. If
a
,
b
,
c
are in arithmetic progression, then the locus of P is
‌
|
√
2
y
|
=
2
|
x
−
y
+
2
|
−
|
x
+
y
−
2
|
‌
√
2
|
y
|
=
|
x
−
y
+
2
|
−
|
x
+
y
−
2
|
‌
2
|
x
−
y
+
2
|
=
|
‌
x
+
y
−
2
√
2
|
+
|
‌
x
−
y
−
2
√
2
|
‌
2
|
x
−
y
+
2
|
=
|
x
+
(
√
2
+
1
)
y
+
2
|
Validate
Solution:
Equation of
A
B
:
y
−
2
=
−
1
(
x
)
Equation of
B
C
:
y
−
2
=
1
(
x
)
Equation of
A
C
:
y
−
0
=
0
(
x
−
2
)
Let the variable point be
(
x
,
y
)
.
a
=
‌
|
x
+
y
−
2
|
√
2
,
b
=
‌
|
x
−
y
+
2
|
√
2
c
=
|
y
|
2
b
=
a
+
c
2
|
‌
x
−
y
+
2
√
2
|
=
|
‌
x
+
y
−
2
√
2
|
+
|
y
|
Hence lows is
|
√
2
y
|
=
2
|
x
−
y
+
2
|
−
|
x
+
y
−
2
|
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