# Which are the points from plane xoy for the relation x^2 - y^2 = x-y, to be true ?

hala718 | Certified Educator

x^2 - y^2 = x-y

==> x^2 - y^2 -x + y = 0

But x^2 - y^2 = (x-y)(x+y

==> (x-y)(x+y) -(x-y) = 0

Now take (x-y) as a common factor:

==> (x-y)[(x+y)-1]= 0

==> (x-y)(x+y-1) =0

==> x+ y -1= 0

==> y= 1-x    and y=x   are the points for the plane.

giorgiana1976 | Student

Let’s analyze the relationship given in enunciation:

x^2 – y^2=x-y

Let's express the difference of squares, x^2 – y^2 =( x-y)*(x+y)

Let’s substitute the difference of squares with the specific product:

( x-y)*(x+y)=(x-y)

We’ll subtract (x-y) both sides:

( x-y)*(x+y)- (x-y)=0

We'll factorize by (x-y).

(x-y)*(x+y-1)=0

From the relation above results:

x-y=0, so x=y

or x+y-1=0. In this case, the points which belong to the line y=1-x are the points for the relation x^2 – y^2=x-y to be true.

neela | Student

To determine points that satisfy x^2-y^2 =  (x-y)

Solution:

x^2-y^2 = x-y implies

x^2-y^2 - (x-y) = 0

(x+y)(x-y) - (x-y) = 0

(x-y){x+y-1} = 0

Therefore all the points in XOY plane that are on the lines

x-y = 0 , Or  x+y-1 = 0

satisfies the realtion x^2-y^2 = x-y.