Question

find the point on the X-axis which is equidistant from A(-3,4 and B(1,-4).
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Answer 1

Answer:

The answer is $(-1,0)$

Step-by-step explanation:

Let the required point on x axis is $P(x,0)$

Given it is equidistant from A(-3,4) and B(1,-4)

Thus using distance formula

                          $AP=BP$

              $\Rightarrow \sqrt{(x+3)^2+(0-4)^2}=\sqrt{(x-1)^2+(0+4)^2}$

Square both sides

             $\Rightarrow (x+3)^2+4^2=(x-1)^2+4^2\\\Rightarrow x^2+6x+9=x^2-2x+1\\\Rightarrow 6x+2x=1-9\\\Rightarrow 8x=-8\\\Rightarrow x=-1$

Answer 2

Answer:

Step-by-step explanation:

Answer:

LET THE POINT ON THE X-AXIS BE (X,0)

BY DISTANCE FORMULA

$\sqrt{(X+3)^{2} +(-4)^{2} }$ = $\sqrt{(X-1)^{2}+(4)^{2} }$

(X+3)^2 +16 = (X-1)^2 + 16

ON SOLVING FURTHER WE GET,

9+6X=1-2X

8X=-8

X=-1

POINT IS (-1,0)

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Step-by-step explanation: