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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?

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1 Answer 1

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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.

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  • $\begingroup$ $ -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 $\endgroup$
    – Jdeep
    May 22, 2020 at 5:04

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