A point P (2, -1) is equidistant from the
points (a, 7) and (-3, a). Find a.
Answers
Answered by
2
Step-by-step explanation:
We know that the distance between the two points is given by the Distance Formula,
√
(
x
1
−
x
2
)
2
+
(
y
1
−
y
2
)
2
(
1
)
To find the distance between
P
A
, substitute the values of
P
(
x
,
0
)
and
A
(
2
,
−
5
)
in Equation (1),
=
√
(
x
−
2
)
2
+
(
0
−
(
−
5
)
)
2
=
√
(
x
−
2
)
2
+
(
5
)
2
To find the distance between
P
B
, substitute the values of
P
(
x
,
0
)
and
B
(
−
2
,
9
)
in Equation (1),
Distance
=
√
(
x
−
(
−
2
)
)
2
+
(
0
−
(
−
9
)
)
2
=
√
(
x
+
2
)
2
+
(
9
)
2
By the given condition, these distances are equal in measure.
Hence
P
A
=
P
B
√
(
x
−
2
)
2
+
(
5
)
2
=
√
(
x
+
2
)
2
+
(
9
)
2
Squaring on both sides
(
x
−
2
)
2
+
25
=
(
x
+
2
)
2
+
81
x
2
+
4
−
4
x
+
25
=
x
2
+
4
+
4
x
+
81
8
x
=
25
−
81
8
x
=
−
56
x
=
−
7
Therefore, the point equidistant from the given points on the axis is
(
−
7
,
0
)
.
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