Question

15. Find a relation between x and y such that the point (x,y) is equidistant from the point
(-2,8) and (-3,-5)​

Answer 1

Step-by-step explanation:

Points (−2,8) and (−3,−5).

Let (x,y) be x

Let (−2,8)=A(−3,−5)=B

AX=BX (length)

AX

2

=BX

2

By Distance formula

(x+2)

2

+(y−8)

2

=(x+3)

2

+(y+5)

2

x

2

+4x+4+y

2

−16y+64=

x

2

+6x+9+y

2

+10y+25

4x+4−16y+64=6x+9+10y+25

4x−16y+68=6x+10y+34

Put x and y in RHS

34=2x+26y

⇒34=2x+26y

⇒17=x+12y

This is a reaction of x and y which will follow the Pt. (x,y) if equidistant from A and B.

Answer 2

Step-by-step explanation:Step-by-step explanation:

Points (−2,8) and (−3,−5).

Let (x,y) be x

Let (−2,8)=A(−3,−5)=B

AX=BX (length)

AX

2

=BX

2

By Distance formula

(x+2)

2

+(y−8)

2

=(x+3)

2

+(y+5)

2

x

2

+4x+4+y

2

−16y+64=

x

2

+6x+9+y

2

+10y+25

4x+4−16y+64=6x+9+10y+25

4x−16y+68=6x+10y+34

Put x and y in RHS

34=2x+26y

⇒34=2x+26y

⇒17=x+12y

This is a reaction of x and y which will follow the Pt. (x,y) if equidistant from A and B.