.....,......see in picture
Attachments:
![](https://hi-static.z-dn.net/files/d26/2682bed356ac86a99d9a33bb6a8591b9.jpg)
sharpyy:
ek baar xy points ka 1 point se distance nikaalo
Answers
Answered by
5
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.
Similar questions
Computer Science,
9 months ago
Science,
9 months ago
Math,
1 year ago
Math,
1 year ago
Biology,
1 year ago