A point p(7,a) is equidistant from the points (2,-9) and(20,a) find a
Answers
Answer:
We know that the distance between the two points (x
1
,y
1
) and (x
2
,y
2
) is
d=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
Let the given points be A=(a,7) and B=(−3,a) and the third point given is P(2,−1).
We first find the distance between P(2,−1) and A=(a,7) as follows:
PA=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(a−2)
2
+(7−(−1))
2
=
(a−2)
2
+(7+1)
2
=
(a−2)
2
+8
2
=
(a−2)
2
+64
Similarly, the distance between P(2,−1) and B=(−3,a) is:
PB=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(−3−2)
2
+(a−(−1))
2
=
(−5)
2
+(a+1)
2
=
25+(a+1)
2
Since the point P(2,−1) is equidistant from the points A(a,7) and B=(−3,a), therefore, PA=PB that is:
(a−2)
2
+64
=
25+(a+1)
2
⇒(
(a−2)
2
+64
)
2
=(
25+(a+1)
2
)
2
⇒(a−2)
2
+64=25+(a+1)
2
⇒(a−2)
2
−(a+1)
2
=25−64
⇒(a
2
+4−4a)−(a
2
+1+2a)=−39(∵(a−b)
2
=a
2
+b
2
−2ab,(a+b)
2
=a
2
+b
2
+2ab)
⇒a
2
+4−4a−a
2
−1−2a=−39
⇒−6a+3=−39
⇒−6a=−39−3
⇒−6a=−42
⇒a=
−6
−42
=7
Hence, a=7.
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