# Given that B has rational co-ordinates and A , B are equidistant from H. I need to find B(h,k).

I am given two points $$A(3,5)$$ and $$H (\sqrt{2},\sqrt{5})$$ and a third point $$B(h,k)$$ . Given that $$B$$ has rational co-ordinates and A , B are equidistant from H. I need to find $$B(h,k)$$.

What I tried :

$$AH^2 = BH^2$$

$$h^2 + k^2 - 2\sqrt{2}(h-3) - 2\sqrt{5}(k-5)-34=0$$ ------------>(1)

In the book , it is given that since $$B(h,k)$$ are rational so $$h-3=0 , k-5=0$$

I didn't understand this step . Eq(1) gives the locus of possible points B. So what does $$h-3=0$$ and $$k-5=0$$ got to do with $$(h,k)$$ being rational?

The sum of a (nonzero) rational number and an irrational number is never zero. Therefore, the expression labeled as equation one is only zero if the irrational terms are each themselves zero. This yields $$2\sqrt{2}(h-3)=0$$ and $$-2\sqrt{5}(k-5)=0$$, from which the result follows.
• $-2\sqrt{2}(h-3)$ , $- 2\sqrt{5}(k-5)$ will always be irrational for all $(h,k)$ except when $(h,k=3,5)$ . But Eq(1) is of a circle so there will be infinte $(h,k)$ that satisfies the equation. This contradicts your theory that sum of a non zero rational number and an irrational number is never zero May 22, 2020 at 5:04