, find the point on the x axis which is equidistant from (2, -5)and(-2

1.	, find the point on the x axis which is equidistant from (2, -5)and(-2, 9)
Answer» We know that a point on the x-axis is of the form (x, 0). So, let the point P(x, 0) be equidistant from A(2, –5) and B(–2, 9). ThenPA = PB{tex}\\Rightarrow{/tex} PA2 = PB2{tex}\\Rightarrow{/tex} (2 - x)2 + (-5 - 0)2 = (-2 - x)2 + (9 - 0)2{tex}\\Rightarrow{/tex} 4 + x2 - 4x + 25 = 4 + x2 + 4x + 81{tex}\\Rightarrow{/tex} - 4x + 25 = 4x + 81{tex}\\Rightarrow{/tex} 8x = -56{tex}\\Rightarrow \\;x = \\frac{{ - 56}}{8} = - 7{/tex}Hence, the required point is (-7, 0)Check:{tex}PA = \\sqrt {{{\\{ 2 - ( - 7)\\} }^2} + {{( - 5 - 0)}^2}}{/tex}{tex}= \\sqrt {81 + 25} = \\sqrt {106}{/tex}{tex}PB = \\sqrt {{{\\{ - 2 - ( - 7)\\} }^2} + {{(9 - 0)}^2}}{/tex}{tex}= \\sqrt {25 - 81} = \\sqrt {106}{/tex}{tex}\\because{/tex} PA = PB{tex}\\therefore{/tex} Our solution is chekcked.

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