if point 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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Answered by
4
Distace between the points (x, y) and (a+b, b-a) & (a-b, a+b) is equal
⇒ √{[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(a + b)
⇒ -2ax - 2bx - 2by + 2ay = - 2ax + 2bx - 2ay - 2by
⇒ ay - bx = bx - ay
⇒ 2ay = 2bx
⇒ bx = ay
Hence proved.
⇒ √{[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(a + b)
⇒ -2ax - 2bx - 2by + 2ay = - 2ax + 2bx - 2ay - 2by
⇒ ay - bx = bx - ay
⇒ 2ay = 2bx
⇒ bx = ay
Hence proved.
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Answered by
14
As Point P be equidistant from A and B
Then, AP = BP
So, we know that
Distance = √(x₂-x₁)² - (y₂-y₁)²
Let us square on both sides in AP = BP
AP² = BP²
And so the roots get cancelled
Then,
(x-a-b)² + (y+a-b)² = (x-a+b)² + (y-a-b)²
x² + a² + b² - 2ax + 2ab - 2bx + y² + a² + b² + 2ay - 2ab -2by
= x² + a² + b² -2ax + 2bx - 2ab + y² + a² + b² - 2ay + 2ab - 2by
(From Formula (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ac )
2ay - 2bx = 2bx - 2ay
(cancelling the opp sides therms as it becomes opp signs)
2ay + 2ay = 2bx + 2bx
4ay = 4bx
ay = bx
(cancelling 4 on both sides)
bx = ay
HENCE PROVED
☺ Hope this Helps ☺
Then, AP = BP
So, we know that
Distance = √(x₂-x₁)² - (y₂-y₁)²
Let us square on both sides in AP = BP
AP² = BP²
And so the roots get cancelled
Then,
(x-a-b)² + (y+a-b)² = (x-a+b)² + (y-a-b)²
x² + a² + b² - 2ax + 2ab - 2bx + y² + a² + b² + 2ay - 2ab -2by
= x² + a² + b² -2ax + 2bx - 2ab + y² + a² + b² - 2ay + 2ab - 2by
(From Formula (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ac )
2ay - 2bx = 2bx - 2ay
(cancelling the opp sides therms as it becomes opp signs)
2ay + 2ay = 2bx + 2bx
4ay = 4bx
ay = bx
(cancelling 4 on both sides)
bx = ay
HENCE PROVED
☺ Hope this Helps ☺
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