co ordinate angry

[sumit saurav]

getting the same point

If the point (x,y) be equidistant from the point A ( a+b,a-b) and B (a-b, a+b) then:
a)ax=by
b) bx = ay
after solving i get a+b=a-b
meaning b=0 so both points turn out to be same

then what is the purpose of this question?

 

[pka]

If the point (x,y) be equidistant from the point A ( a+b,a-b) and B (a-bb, a+b) then:
a)ax=by
b) bx = ay

If \(\displaystyle P(p,q)~\;\& \;S~(s,t)\) are two points then every point on the line \(\displaystyle \displaystyle\left( {y - \frac{{q + t}}{2}} \right) = \frac{{s - p}}{{q - t}}\left( {x - \frac{{p + s}}{2}} \right)\) is equally distant from \(\displaystyle P~\&~S\).

 

[sumit saurav]

by applying the suggested method i got
x=y
which doesnt lead to either of the options

 

[mmm4444bot]

by applying the suggested method i got
x=y
which doesnt lead to either of the options

Maybe you made a mistake. We cannot see what you did.

Please start showing your work.

Cheers :cool:

 

[sumit saurav]

since equidistant
(a+b-x)square +(a-b-y)square=(a-b-x)square+(a+b -y)square
=-2x(a+b) -2y(a-b)= -2x(a-b) -2y(a+b)
=x(a+b)-x(a-b)=y(a+b)-y(a-b)
=x=y
it doesnt help but the process also doesnt seem wrong

 

[lookagain]

If the point (x,y) be equidistant from the point A ( a+b,a-b)
and B (a-bb, a+b) <--- You have a typo here. The first co-ordinate should be just "a-b."
then:
a)ax=by
b) bx = ay
after solving i get a+b=a-b
meaning b=0 so both points turn out to be same

.

 

[sumit saurav]

i corrected the spell mistake but it doesnt help in getting the answer