Question

if the point a(2,-5)is equidistant from p(4,5)and (-12,y), find the value of y.​

Answer 1

Answer:

Since A(3,y) is equidistant from points P(8,−3) and Q(7,6), then,

AP=AQ

(3−8)

2

+(y−(−3))

2

=

(3−7)

2

+(y−6)

2

(5)

2

+(y+3)

2

=

(4)

2

+(y−6)

2

25+y

2

+9+6y

=

16+y

2

+36−12y

y

2

+6y+34

=

y

2

−12y+52

Squaring both sides,

y

2

+6y+34=y

2

−12y+52

18y=18

y=1

AQ=

(3−7)

2

+(1−6)

2

=

(4)

2

+(5)

2

=

16+25