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the equation of the locus of a point equidistant f
Question:
The equation of the locus of a point equidistant from the point A(1, 3) and B(-2, 1) is:
MHT CET
Updated On:
Jun 23, 2024
(A)
6
x
−
4
y
=
5
(B)
6
x
+
4
y
=
5
(C)
6
x
+
4
y
=
7
(D)
6
x
−
4
y
=
7
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The Correct Option is
B
Solution and Explanation
Explanation:
Let
P
(
h
,
k
)
be any point on the locus which is equidistant from the point A(1, 3) and B(-2, 1).According to question,
P
A
=
P
B
(
h
−
1
)
2
+
(
k
−
3
)
2
=
(
h
+
2
)
2
+
(
k
−
1
)
2
..(i)Squaring both side in equation (i) we get,
⇒
(
h
−
1
)
2
+
(
k
−
3
)
2
=
(
h
+
2
)
2
+
(
k
−
1
)
2
(
∵
(
a
±
b
)
2
=
a
2
+
b
2
±
2
a
b
)
⇒
h
2
−
2
h
+
1
+
k
2
−
6
k
+
9
=
h
2
+
4
h
+
4
+
k
2
−
2
k
+
1
⇒
−
2
h
−
6
k
+
10
=
4
h
−
2
k
+
5
⇒
6
h
+
4
k
=
5
∴
The locus of
(
h
,
k
)
is
6
x
+
4
y
=
5
.Hence, the correct option is (B).
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