Question

find the point on X -axis which is equidistance from (2,6) and (-4,8).
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Answer 1

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

(X,0) POINT IS equidistance from (2,6) and (-4,8).

APPLYING THE DISTANCE FORMULA , WE GET ,

$\sqrt{ {(x - 2)}^{2} + {(0 - 6)}^{2} } = \sqrt{ {(x + 4)}^{2} + {(0 - 8)}^{2} } \\ = > {(x - 2)}^{2} + 36 = {(x + 4)}^{2} + 64 \\ = > {(x - 2)}^{2} - {(x + 4)}^{2} = 64 - 36 \\ = > {x}^{2} - 2x + 4 - {x}^{2} - 8x + 16 = 28 \\ = > - 10x + 20 = 28 \\ = > - 10x = 8 \\ = > x = \frac{ - 8}{10} \\ = > x = \frac{ - 4}{5}$

SO THE POINT ON X AXIS IS (-4/5, 0)

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Answer 2

Given points:-

• (2, 6) and (-4, 8)

To find:-

• The point on x-axis which is equidistant from the given points.

Solution:-

Let us name the given points.

Let A and B be the points on (2, 6) and (-4, 8) respectively.

Now,

As we need to find the point on x-axis, the point on x-axis = (x,0)

Let us name this point with C

Now,

We are given that these points are equidistant from x-axis.

Hence,

AC = BC ................. (i)

Now,

Let us find the distance between AC and BC,

For AC,

The points are as follows:-

• A(2, 6) and C(x, 0)

Hence,

$\sf{x_1 = 2\:and\:x_2 = x}$

$\sf{y_1 = 6\:and\:y_2 = 0}$

The distance formula is as follows:-

$\sf{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$

Putting respective values:-

$\sf{AC = \sqrt{(x - 2)^2 + (0 - 6)^2}}$

Using identity (a - b)² = a² + b² - 2ab

$\sf{\implies AC = \sqrt{(x^2 + 4 - 4x) + (-6)^2}}$

$\sf{\implies AC = \sqrt{x^2 + 4 - 4x + 36}}$

$\sf{\implies AC =\sqrt{x^2 - 4x + 40}.............(ii)}$

For BC,

The points are as follows:-

• B(-4, 8) and C(x, 0)

Here,

$\sf{x_1 = -4\:and\:x_2 = x}$

$\sf{y_1 = 8\:and\:y_2 = 0}$

Putting the respective values in the distance formula:-

$\sf{BC = \sqrt{[x - (-4)]^2 + (0 - 8)^2}}$

$\sf{\implies BC = \sqrt{(x + 4)^2 + (-8)^2}}$

Using identity (a + b)² = a² + b² + 2ab

$\sf{\implies BC = \sqrt{[(x)^2 + (4)^2 + 2\times 4\times x] + 64}}$

$\sf{\implies BC = \sqrt{x^2 + 16 + 8x + 64}}$

$\sf{\implies BC = \sqrt{x^2 + 8x + 80}}$

From equation (i) we have,

AC = BC

Putting the respective values from equation (ii) and (iii)

$\sf{\sqrt{x^2 - 4x + 40} = \sqrt{x^2 +8x + 80}}$

Squaring both the sides,

$\sf{(\sqrt{x^2 - 4x + 40})^2 = (\sqrt{x^2 + 8x + 80)^2}}$

= $\sf{x^2 - 4x + 40 = x^2 + 8x + 80}$

= $\sf{\not{x^2} - 4x + 40 = \not{x^2} + 8x + 80}$

= $\sf{-4x + 40 = 8x + 80}$

= $\sf{-8x - 4x = 80 - 40}$

= $\sf{-12x = 40}$

=> $\sf{x = \dfrac{40}{-12}}$

=> $\sf{x = \dfrac{-10}{3}}$

Therefore the point on x-axis which is equidistant to points (2, 6) and (-4, 8) is (-10/3, 0)

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