Question

if Q(0,1)is equidistant form P(5,-3) and R(X,6),fined the value of X
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Answer 1

Given :

• Q(0,1) is equidistant from P(5,-3) & R(X,6)

To find : X

Formula used :

Distance = √ { (x₂ - x₁)² + (y₂ - y₁² }

Solution :

Let distance between P(5,-3) & Q(0,1) = A

then,

$\sf \implies \: A \: = \sqrt{ {(5 - 0)}^{2} + {( - 3 - 1)}^{2} } \\ \\ \sf \implies \: A \: = \: \sqrt{ {5}^{2} + {( - 4)}^{2} } \\ \\ \sf \implies \: A \: = \: \sqrt{25 + 16} \\ \\ \sf \implies \: A \: = \sqrt{41}$

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Let , distance between R(X,6) & Q(0,1) = B

$\sf \implies \: B \: = \sqrt{ {( X - 0) }^{2} + {(6 - 1)}^{2} } \\ \\ \sf \implies \: B \: = \sqrt{ {( X ) }^{2} + {(5)}^{2} } \\ \\ \sf \implies \: B \: = \sqrt{ {X }^{2} + 25 }$

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As , Q(0,1) is equidistant from P(5,-3) & R(X,6)

➝ A = B

➝ √41 = √(X² + 25)

On Squaring both side, we get;

➝ 41 = X² + 25

➝ X² = 41 - 25

➝ X² = 16

➝ X = √16

➝ X = ±4

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ANSWER :

X = ±4