Question

The point on the x-axis which is equidistant from (- 4, 0) and (10, 0)

Answer 1

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Find the point on the x-axis which is equidistant from (- 4, 0) and (10, 0).

$\huge\underline\mathbb{\red S\pink{O}\purple{LU} \blue{T} \orange{IO}\green{N :}}$

★ Given that,

• A(- 4, 0)
• B(10, 0)

★ Let,

◼ " P " is a point which lies on the X - axis.

Since, the co - ordinate is P(x, 0).

◼ Hence, the point P is equidistant from the points A & B.

◼ Therefore distance PA = PB.

Formula for distance is

$\tt\:⟹ \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^2}</p><p>$

For the points PA = P(x, 0) & A(- 4, 0)

• x1 = x ; y1 = 0
• x2 = - 4 ; y2 = 0

$\tt\:⟹PA = \sqrt{( - 4 - x)² + (0 - 0)²}$

$\tt\:⟹ PA = \sqrt{ {(4 - x)}^{2} }$

For the points PB = P(x, 0) & B(10, 0)

• x1 = x ; y1 = 0
• x2 = 10 ; y2 = 0

$\tt\:⟹PB = \sqrt{ {(10 - x)}^{2} + {(0 - 0)}^{2} }$

$\tt\:⟹ PB = \sqrt{ {(10 - x)}^{2} }$

➡ PA = PB

$\tt\:⟹ \sqrt{ {(4 - x)}^{2} } = \sqrt{ {(10 - x)}^{2} }$

• Squaring on both sides.

$\tt\:⟹ { (\sqrt{ {(4 - x)}^{2} } )}^{2} = {( \sqrt{ {(10 - x)}^{2} } }^{2} )$

$\tt\:⟹ {(4 - x)}^{2} = {(10 - x)}^{2}$

• (a - b)² = a² + b² - 2ab

$\tt\:⟹ 16 + {x}^{2} - 8x = 100 + {x}^{2} - 20x$

$\tt\:⟹16 - 8x = 100 - 20x$

$\tt\:⟹100 - 20x + 8x - 16 = 0$

$\tt\:⟹84 - 12x = 0$

$\tt\:⟹84 = 0 + 12x$

$\tt\:⟹12x = 84$

$\tt\:⟹x = \frac{84}{12}$

$\tt\:⟹x = 7$

$\underline{\boxed{\bf{\purple{∴ Hence, the \: required \: point \: P(x, 0) = P(7, 0)}}}}</p><p></p><p></p><p></p><p></p><p>$

Answer 2

Answer:

Given two point on the x -axis (-4,0) and (10,0).

We have to find the point on the x-axis which is equidistant from the two point.

Let the coordinates of the point be (x,y).

The point which is equidistant from the two points will be their midpoint .

Using midpoint formula -

x = (-4+10)/2 = 3

y = 0

The coordinates of the points are (3,0) . So the option D) is the correct answer.

Step-by-step explanation: