Question

If a point (x, y) is equidistant from the points (a+b, b-a) and (a-b, a+b) then show that ay=bx.

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Answer 1

Step-by-step explanation:

Let (x,y)= point P

Therefore, using mid-point formula

(a+b+a-b)/2=x

2x=2a

x=a.....equation 1

(b-a+a+b)/2=y

2y=2b

y=b.....equation 2

Now in equation 1

x=a

Multiplying both sides by b we get:

xb=ab

From equation 2,y=b therefore we get:

xb=ay

Which is the answer you are looking for:)

Answer 2

Answer:

By Distance Formula

$= \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }$

ATQ:

$\sqrt{ {(a + b - x)}^{2} + {(b - a - y)}^{2} } = \sqrt{ {(a - b - x)}^{2} + {(a + b - y)}^{2} } \\ {(a + b - x)}^{2} + {(b - a - y)}^{2} = {(a - b - x)}^{2} + {(a + b - y)}^{2} \\ {a}^{2} + {b}^{2} + {x}^{2} + 2ab - 2bx - 2ax + {a}^{2} + {b}^{2} + {y}^{2} - 2ab + 2ay \\ - 2by = {a}^{2} + {b}^{2} + {x}^{2} - 2ab + 2bx - 2ax + {a}^{2} + {b}^{2} + {y}^{2} + 2ab \\ - 2by - 2ay \\ - 2bx + 2ay = 2bx - 2ay \\ 4ay = 4bx \\ ay = bx$

Therefore,

$bx = ay$

HOPE THIS ANSWER WILL HELP YOU........

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