Question

find the point on the y-axis which is equidistant from points 5, -2,
and 3,-2

Answer 1

Step-by-step explanation:

Let P(0,y) be any point on y-axis.

Let A(5,−2) and B(−3,2). Then,

PA=PB

PA2=PB2

(5−0)2+(−2−y)2=(−3−0)2+(2−y)2

25+4+y2+4y=9+4+y2−4y

8y=−16

y=−2

Therefore, the required point is (0,−2)

Answer 2

Answer:

P(4,0)

Step-by-step explanation:

if point is on y axis then point is=(x,0)

let consider point P on y Axis

A and B are two points equidistant from that point.

hence, PA=PB. (1)

apply distance formula

d=√(x1-x2)^2+(y1-y2)^2

then we get

p(x,y) and A(5,-2)

d=√(x-5)^2+(y-2)^2. (PA)

P((x,y) and B(3,-2)

d=√(x-3)^2+(y+2)^2. (PB)

PA=PB

squaring on both sides

(x-5)^2+(y+2)^2=(x-3)^2+(y+2)^2

x^2+9-6x+y^2+4+4y=x^2+25-10x+y^2+4+4y

9-6x=25-10x

4x=16

x=4

hence, that point P(x,0)

p(4,0)