Question

Find a relation between r and y such that the point (x, y) is equidistant from the points
(-2.8) and (-3.-5)​

Answer 1

by using formula between (pA)^2= (PA)^2

Step-by-step explanation:

Let the given point on X axis be p(x,0)

( x-+2)^2 +(0+8)^2=(x+3)^2+(0+5)^2

x^2+4-2×2×x+64=x^2 +9-2×3×x+25

4x+60=6x-34

2x=26

x=13

now

Let the givan point on Y axis p(y,0)

y^2+64-16y+4=y^2+25-10y+9

68-16y= 16-10y

6y=84

y=16

heance( 13,16)

Answer 2

halo mate

here is your answer

by using formula between (pA)^2= (PA)^2

X axis be p(x,0)

( x-+2)^2 +(0+8)^2=(x+3)^2+(0+5)^2

x^2+4-2×2×x+64=x^2 +9-2×3×x+25

4x+60=6x-34

2x=26

x=13

now

Y axis p(y,0)

y^2+64-16y+4=y^2+25-10y+9

68-16y= 16-10y

6y=84

y=16

hence( 13,16)