Question

4. Find the point on the y-axis, which is equidistant from the points A (-3, 2) and B (5,-2).​

Answer 1

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=(5,2) and B=(−4,3) and let the point on y-axis be P(0,y).

We first find the distance between P(0,y) and A=(5,2) as follows:

PA=
(x
2
​
−x
1
​
)
2
+(y
2
​
−y
1
​
)
2

​
=
(5−0)
2
+(2−y)
2

​
=
5
2
+(2−y)
2

​
=
25+(2−y)
2

​


Similarly, the distance between P(0,y) and B=(−4,3) is:

PB=
(x
2
​
−x
1
​
)
2
+(y
2
​
−y
1
​
)
2

​
=
(−4−0)
2
+(3−y)
2

​
=
(−4)
2
+(3−y)
2

​
=
16+(3−y)
2

​


Since the point P(0,y) is equidistant from the points A=(5,2) and B=(−4,3), therefore, PA=PB that is:

25+(2−y)
2

​
=
16+(3−y)
2

​


⇒(
25+(2−y)
2

​
)
2
=(
16+(3−y)
2

​
)
2


⇒25+(2−y)
2
=16+(3−y)
2


⇒(2−y)
2
−(3−y)
2
=16−25

⇒(4+y
2
−4y)−(9+y
2
−6y)=−9(∵(a−b)
2
=a
2
+b
2
−2ab)

⇒4+y
2
−4y−9−y
2
+6y=−9

⇒2y−5=−9

⇒2y=−9+5

⇒2y=−4

⇒y=
2
−4
​
=−2

Hence, the point on the y-axis is (0,−2).

Hope it will help you