Question

Find a point on x-axis which is equidistant from the points (5, 5) and (-2, 4).

Answer 1

Answer:

AC=BC (Equidistance)

AC

2

=BC

2

(x−5)

2

+(0−4)

2

=(x+2)

2

+(0−3)

2

x

2

−10x+25+16=x

2

+4+4x+9

−14x+41−13=0

−14x+28=0

=14x=−28

x=28/14

x=2

Step-by-step explanation:

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