Question

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(a) Find x if the distance between (-8, x) and (2, 0) is 5√5

(b) Find the point on the x-axis which is equidistant from (2,-5) and (-2,9)

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Answer 1

Answer:-

a) Given:-

Distance between ( - 8 , x) and (2 , 0) is 5√5 units.

We know that,

Distance between two points (x₁ , y₁) & (x₂ , y₂) is:

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

Let,

• x ₁ = -8

• y₁ = x

• x₂ = 2

• y₂ = 0

So, According to the question,

$\implies \sf \: \sqrt{ {[2 - ( - 8)]}^{2} + (0 - x) ^{2} } = 5 \sqrt{5} \\ \\ \\ \implies \sf \: \sqrt{( {2 + 8)}^{2} + ( - x) ^{2} } = 5 \sqrt{5}$

On squaring both sides we get,

$\implies \sf \: {(10)}^{2} + {x}^{2} = 25 \times 5 \\ \\ \\ \implies \sf \:100 + {x}^{2} = 125 \\ \\ \\ \implies \sf \: {x}^{2} = 125 - 100 \\ \\ \\ \implies \sf \: {x}^{2} = 25 \\ \\ \\ \implies \sf \:x = \sqrt{25} \\ \\ \\ \implies \boxed{\sf \:x = \pm \: 5}$

∴ The value of x is ± 5.

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b)

• If a point is on x - axis , it's y - coordinate is equal to zero.

Let, the x - coordinate be x.

So, the point will be (x , 0).

It is given that:

(x , 0) is equidistant from (2,-5) and (-2,9)

That means;

Distance between (x , 0) and (2 , - 5) = Distance between (x , 0) and (- 2 , 9)

First, we can find the distance between the points seperately and then calculate the coordinates of the point.

Distance between (x , 0) and (2 , - 5) :

Here,

• x ₁ = x

• y₁ = 0

• x₂ = 2

• y₂ = - 5

Hence,

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

Now,

Distance between (x , 0) and (- 2 , 9) :

Here,

• x ₁ = x

• y₁ = 0

• x₂ = - 2

• y₂ = 9

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

Now making them equal we get,

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

Now squaring both sides we get,

$\implies \sf \: (2 - x) ^{2} + 25 = ( - 2 - x) ^{2} + 81$

using (a - b)² = a² + b² - 2ab we get,

$\implies \sf \: {2}^{2} + \cancel{ {x}^{2} } - 2(2)(x) + 25 = {( - 2)}^{2} + \cancel{{x}^{2} } - 2( - 2)(x) + 81 \\ \\ \\ \implies \sf \cancel{ 4} - 4x + 25 = \cancel{4 }+ 4x + 81 \\ \\ \\ \implies \sf \: 25 - 4x = 81 + 4x \\ \\ \\ \implies \sf \: 25 - 81 = 4x + 4x \\ \\ \\ \implies \sf \: - 56 = 8x \\ \\ \\ \implies \sf \frac{ - 56}{8} = x \\ \\ \\ \implies \boxed{\sf \: x = - 7}$

∴ The required point (x , 0) = ( - 7 , 0).

Answer 2

Answer:

$\bf(a) \: \: x = ±5$

$\bf(b) \: x = - 7$

Step-by-step explanation:

$\bf(a)$

$\bf \: Let \: the \: points \: be$

$\bf \: A (-8, x)$

$\bf \: B (2, 0)$

$\bf \: Applying \: distance \: formula,$

$\bf \: AB = √[(x₂ - x₁)² + (y₂ - y₁)²]$

$\bf \: Here:$

$\bf \: x₁ = -8$

$\bf \: x₂ = 2$

$\bf \: y₁ = x$

$\bf \: y₂ = 0$

Substitute these values in the above formula

$\bf= >AB = √[(2 - (-8))² + (0 - x)²]$

$\bf = > AB = √[(2 + 8)² + (-x)²]$

$\bf = > AB = √[(10)² + x²]$

$\bf = > AB = √(100 + x²)$

$\bf= > 5√5 = √(100 + x²)$

$\bf \: Squaring \: on \: both \: sides,$

$\bf (5√5)² = 100 + x²$

$\bf = > 125 = 100 + x²$

$\bf = > 125 - 100 = x²$

$\bf = > x² = 25$

$\bf= > x = √25$

$\bf∴ x = ±5$

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$\bf(b)$

Let the point be (x, 0) on x–axis which is equidistant from (2, –5) and (–2, 9).

Using Distance Formula and according to given conditions we have:

$\bf \sqrt{(x−2)2+(0−(−5))2}$

$\bf= \sqrt{(x−(−2))2+(0−9)2}$

$\bf \: Squaring \: both \: sides, \: we \: get$

$\bf = > x2+4−4x+25=x2+4+4x+81$

$\bf= > −4x+29=4x+85$

$\bf = > 8x=−56$

$\bf = > x=−7$

Therefore, point on the x–axis which is equidistant from (2, –5) and (–2, 9) is (–7, 0)