find the value of y if the distance between point on the Y-axis which is equidistant from AB²=[(-1)-2]² +[y-(-2)]²... ........... 5²=(-3)²+ ............. 25= ................
Answers
Answer:
Step-by-step explanation:
We know that the distance between the two points (x1 , y1 ) and (x2, y2 ) 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= (x2 −x1 )2+(y2−y1 )2 =(5−0) 2 +(2−y) 2 =52 +(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).