Question

(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

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

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Let the points be

A (-8, x)

B (2, 0)

Applying distance formula,

AB = √[(x₂ - x₁)² + (y₂ - y₁)²]

Here:

x₁ = -8

x₂ = 2

y₁ = x

y₂ = 0

Substitute these values in the above formula

⇒ AB = √[(2 - (-8))² + (0 - x)²]

⇒ AB = √[(2 + 8)² + (-x)²]

⇒ AB = √[(10)² + x²]

⇒ AB = √(100 + x²)

⇒ 5√5 = √(100 + x²)

Squaring on both sides,

(5√5)² = 100 + x²

⇒ 125 = 100 + x²

⇒ 125 - 100 = x²

⇒ x² = 25

⇒ x = √25

∴ x = ±5

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

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Let us consider given Points are A(2,-5) and B(-2,9). If Point is on x - axis than y would be 0. Point P is equidistant from the Points A and B. ... Hence, The Required Point is P(-7,0).