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  • Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

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Let the point which is equidistant from A(2,-5)\ and \ B(-2,9) be X(x,0) as it lies on X-axis.

Then, we have

Distance AX: = \sqrt{(x-2)^2+(0+5)^2}

and Distance BX = \sqrt{(x+2)^2+(0+9)^2}

According to the question, these distances are equal length.

Hence we have,

\sqrt{(x-2)^2+(0+5)^2}   = \sqrt{(x+2)^2+(0+9)^2}

Solving this to get the required coordinates.

Squaring both sides we get,

(x-2)^2+25 = (x+2)^2+81

\Rightarrow (x-2+x+2)(x-2-x-2)= 81 - 25 = 56

\Rightarrow (-8x)= 56

Or, \Rightarrow x = -7

Hence the point is X(-7,0).

Posted by

Divya Prakash Singh

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