Question

Find a point on x-axis wich is equidistant from point A(-1,2) and B(5,4)​

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

Answer 2

Answer:

(3,0)

Step-by-step explanation:

Let that point which is equidistant from points A and B and on x-axis be C(x,0).

Then, AC=BC.

=> (AC)^2 = (BC)^2.

=> {x-(-1)}^2 + (0-2)^2 = (x-5)^2 + (0-4)^2.

=> (x+1)^2 + (-2)^2 = x^2+25-10x +16.

=> x^2+1+2x +4 = x^2+41-10x.

=> 12x = 36.

=> x = 3.

Hence, C (3,0) is the required point.

Hope this helps you.