Question

If the points p(x, y) is equidistant from the points A(a+b, b-a) and B(a-b, a+b).Prove that bx=ay

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Answer 1

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Distace between the points (x, y) and  (a+b, b-a) & (a-b, a+b) is equal  
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⇒ √{[x - (a + b)]2 + [y - (b -a)]2}
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= √{x - (a - b)]2 + [y - (a + b)]2}

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 ⇒ x2 + (a + b)2 - 2x(a + b) + y2 + (b - a)2 - 2y(b - a) = x2 + (a - b)2 - 2x(a - b) + y2 + (a + b)2 - 2y(a + b) 

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⇒ -2ax - 2bx - 2by + 2ay = - 2ax + 2bx - 2ay - 2by 
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⇒ ay - bx = bx - ay
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 ⇒ 2ay = 2bx 
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⇒ bx = ay$<b><font face=copper black size=4 color=green>$


 Hence proved.

Answer 2

Answer:

bx = ay

Step-by-step explanation:

Since,P(x,y) is equidistant from the A(a + b, b - a) and B(a - b, a + b).

∴ PA = PB

Distance formula:

⇒ √(x - (a + b))² + (y - (b - a))² = √(x - (a - b))² + (y - (a + b))²

On squaring both sides, we get

⇒ (x - (a + b))² + (y - (b - a))² = (x - (a - b))² + (y - (a + b))²

⇒ x² + (a + b)² - 2x(a + b) + y² + (b - a)² - 2y(b - a) = x² + (a - b)² - 2x(a - b) + y² + (a + b)² - 2y(b + a)

⇒ -2x(a + b) - 2y(b - a) = -2x(a - b) - 2y(b + a)

⇒ -2ax - 2bx - 2by + 2ay = -2ax + 2bx - 2by - 2ay

⇒ -2ax - 2bx - 2by + 2ay + 2ax - 2bx + 2by + 2ay = 0

⇒ -4bx + 4ay = 0

⇒ -4(bx - ay) = 0

⇒ bx - ay = 0

⇒ bx = ay.

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