Question

Find the point on x-axis which is equidistant from the points (2,-2) and (-4,2)
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Answer 1

Answer:

(-1, 0)

Step-by-step explanation:

Let the point on x-axis M(x,0)

As it is equidistant from the points A(2,-2) and B(-4,2)

∴ AM = BM

∴ AM² = BM²

∴ (x-2)² + (0 + 2)² = (x + 4)² + (0 - 2)²

x² - 4x +4 + 4 = x² + 8x + 16 + 4

-4x + 8 = 8x + 20

12x = -12

∴ x = -1

∴ the point is having coordinates (-1, 0)

Answer 2

☯ Let the given points be A (2,-2) and B (-4,2) is equidistant from a point P.

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Given that,

• The points are equidistant from x - axis.

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$\therefore$ It's y - coordinate will be 0.

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

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

• Point P is equidistant from A & 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,

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$:\implies\sf \sqrt{(x - 2)^2 + (0 - (-2))^2} = \sqrt{(x - (-4))^2 + (0 - 2)^2}$

$:\implies\sf \bigg(\sqrt{(x - 2)^2 + (0 - (-2))^2}\bigg)^2 = \bigg(\sqrt{(x - (-4))^2 + (0 - 2)^2}\bigg)^2$

$:\implies\sf (x - 2)^2 + (0 + 2)^2 = (x + 4)^2 + (0 - 2)^2\\ \\ \\ :\implies\sf (x - 2)^2 + (2)^2 = (x + 4)^2 + (2)^2\\ \\ \\ :\implies\sf x^2 + 2^2 - 4x + 4 = x^2 + 4^2 + 8x + 4\\ \\ \\ :\implies\sf x^2 + 4 - 4x + 4 = x^2 + 16 + 8x + 4\\ \\ \\ :\implies\sf \cancel{x^2} - 4x + 8 = \cancel{x^2} + 8x + 20\\ \\ \\ :\implies\sf - 4x + 8 = 8x + 20\\ \\ \\ :\implies\sf - 4x - 8x = 20 - 8\\ \\ \\ :\implies\sf - 12x = 12\\ \\ \\ :\implies\sf x = - \cancel{ \dfrac{12}{12}}\\ \\ \\ :\implies{\underline{\boxed{\frak{\purple{x = - 1}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Hence,\:Required\:point\:is\: {\textsf{\textbf{(-1,0)}}}.}}}$