Question

find the point on x axis which eqidistance from (2, -5) and (-2, 9)​

Answer 1

Explanation:

Let (a, 0) be the point on the x axis that is equidistant from the points (2, -5) and (-2, 9).

According to the question,

$\leadsto \tt \: \sqrt{(a - 2) {}^{2} + ( 0 + 5) {}^{2} } = \sqrt{(a + 2) {}^{2} + (0-9) {}^{2} }$

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$\leadsto \tt \: \sqrt{a {}^{2} - 4a + 4 + ( - 5) {}^{2} } = \sqrt{a {}^{2} + 4a + 4 + (9) {}^{2} }$

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$\leadsto \tt \: \sqrt{a {}^{2} - 4a + 4 + 25 } = \sqrt{a {}^{2} + 4a + 4 + 81 }$

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$\tt Squaring \:on \:both \: side, \: we \: get ;$

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$\leadsto \tt \: \cancel{a {}^{2} } - 4a + 29 = \cancel{a {}^{2} } + 4a + 85$

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$\leadsto \tt \: - 4a - 4a = 85 - 29$

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$\leadsto \tt \: - 8a = 56$

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$\leadsto \tt \: a = \cancel{- \dfrac{56}{ 8}}$

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$\leadsto \underline{\boxed{ \tt a = - 7}} \: \bigstar$

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Hence, the required point on x - axis is (-7,0).

Answer 2

☯ Let the given point be A (2,-5) and B (-2,9) is equidistant from a point P.

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Given that,

• Thd points are equidistant from x - axis.

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$\therefore$ Its y - coordinate will be 0.

Therefore, Coordinates of P is (x,0).

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$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

• Point P is equidistant from A and B

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$:\implies\sf AP = BP$

$\dag\;{\underline{\frak{Using\:Distance\:formula,}}}$

$\star\;{\boxed{\sf{\pink{d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}}}}$

Therefore,

$:\implies\sf \sqrt{(x - 2)^2 + (0 - (-5))^2} = \sqrt{(x - (-2))^2 + (0 - 9)^2}\\ \\ \\ :\implies\sf \bigg( \sqrt{(x - 2)^2 + (0 - (-5))^2} \bigg)^2 = \bigg( \sqrt{(x - (-2))^2 + (0 - 9)^2} \bigg)^2$

$:\implies\sf (x - 2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0 - 9)^2\\ \\ \\ :\implies\sf x^2 + 4 - 4x + 25 = x^2 + 4 + 4x + 81\\ \\ \\ :\implies\sf x^2 - 4x + 29 = x^2 + 4x + 85\\ \\ \\ :\implies\sf - 4x + 29 = 4x + 85\\ \\ \\ :\implies\sf - 4x - 4x = 85 - 29\\ \\ \\ :\implies\sf - 8x = 56\\ \\ \\ :\implies\sf x = -\cancel{\dfrac{56}{8}}\\ \\ \\ :\implies{\underline{\boxed{\frak{\purple{x = -7}}}}}\;\bigstar$

$:\therefore\:{\underline{\sf{Hence,\:the\:required\:point\:on\;x-axis\:is\: {\textsf{\textbf{(-7,0)}}}.}}}$