Question

Find a relation blur x and y such that
y such that the point
x, y is equidistant from the point (71) and 2,5)

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Answer 1

Given:

• (x,y) be equidistant from the points (7,1) and (2,5).

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To find:

• Relation between x and y?

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Formula used:

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

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Solution:

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☯ Let P(x,y) be equidistant from the points A(7,1) and B(2,5).

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Therefore,

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Distance of AP = Distance of BP

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Also,

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$:\implies\sf (AP)^2 = (BP)^2$

$:\implies\sf \sqrt{(x - 7)^2 + (y - 1)^2} = \sqrt{(x - 2)^2 + (y - 5)^2}$

$:\implies\sf (x - 7)^2 + (y - 1)^2 = (x - 2)^2 + (y - 5)^2$

$:\implies\sf x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 4x + 4 + y^2 - 10y + 25$

$:\implies\sf - 14x + 49 - 2y + 1 = - 4x + 4 - 10y + 25$

$:\implies\sf - 14x - 2y + 50 = - 4x - 10y + 29$

$:\implies\sf - 14x + 4x - 2y + 10y = 29 - 50$

$:\implies\sf \purple{- 10x + 8y = - 21}$

Answer 2

Answer:

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Step-by-step explanation:

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