Question

find the point on x axis which is equidistant from point (-1,0) and (5,0)

Answer 1

Answer:Let x Axis be p(x,0)

Be equidistant from (-1,0)and(5,0)

AP=BP

AP^2=BP^2

(x+1)^2+(0-0)^2=(x-5)^2+(0-0)^2

x^2+1+2x=x^2+25-10x

X^2-x^2+10x+2x=25-1

12x=24

x=24/12

x=2

Therefore,p(2,0)

Answer 2

Answer:Let x Axis be p(x,0)

Be equidistant from (-1,0)and(5,0)

AP=BP

AP^2=BP^2

(x+1)^2+(0-0)^2=(x-5)^2+(0-0)^2

x^2+1+2x=x^2+25-10x

X^2-x^2+10x+2x=25-1

12x=24

x=24/12

x=2

Therefore,p(2,0)