Question

find the point on the x-axis which is equi distant from (3,5) and (-2,-4)​

Answer 1

Answer:

(2,0)

AC=BC (Equidistance)

AC2=BC2

(x−5)2+(0−4)2=(x+2)2+(0−3)2

x2−10x+25+16=x2+4+4x+9

−14x+41−13=0

−14x+28=0

=14x=−28

x=28/14x=2

Answer 2

Answer:

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