# $ax+by=x^2+y^2\implies a=x$ and $b=y$

I currently edit curriculum for high school geometry and I came across a mistake in one of their diagrams. After doing some work, I boiled down their mistake to an assumption that if $$ax+by=x^2+y^2$$ for $$a,b,x,y>0$$, then $$a$$ is not necessarily equal to $$x$$ and $$b$$ is not necessarily equal to $$y$$. However, after analyzing the corresponding graph of this equation (and using some common sense), I am quite confident that we must have $$a=x$$ and $$b=y$$. Does anyone have any suggestions on how I could show this algebraically? Thank you.

• $$x^2 + y^2 = x^2+xy +y^2-xy = (x+y)x + (y-x)y...$$ – peterwhy Apr 15 at 0:42
• Is it supposed to be true for any value of $x,y$? Usually $a,b$ would be constants, not allowed to be equal to $x,y$. Please show the specific problem. – Ross Millikan Apr 15 at 0:42

Counter example: $$x=1, y=2, a=3, b=1$$.

More generally,

\begin{align*} x^2 +y^2 &= x^2 + txy + y^2 - txy\\ &= (x+ty)x + (y-tx)y \end{align*}

By varying $$t\ne 0$$ within the limit that $$x+ty > 0$$ and $$y-tx>0$$, it is easy to generate counter examples.

It's not true. $$ax+by=c$$ is the equation of a line. At the same time $$x^2+y^2=c$$ is the equation of a circle, centered on origin. From the information, it's only a quarter of it ($$x,y>0$$). What you have is an intersection of a random line with this quarter circle. In the absence of more information, I can get a set of numbers that satisfy this equation, without $$x=a$$ or $$b=y$$

Sorry everyone, I realized I overlooked something silly. Here is (basically) what the problem looked like:

There is a large right triangle and two smaller right triangles whose legs intersect the hypotenuse of the large triangle at its midpoint. I wanted to create conditions on $$a,b,x,y$$ so that their figure was accurate.

Originally, the curriculum had $$a=7,b=22,x=17,$$ and $$y=6$$. I determined this was impossible because $$hyp$$(small triangle)+$$hyp$$(other small triangle) $$\not=$$ $$hyp$$(big triangle). In other words, $$\sqrt{7^2+22^2}+\sqrt{17^2+6^2}\not=\sqrt{(7+17)^2+(22+6)^2}$$.

That is when I decided to set $$\sqrt{a^2+b^2}+\sqrt{x^2+y^2}=\sqrt{(a+x)^2+(b+y)^2}$$ which yielded $$ax+by=x^2+y^2.$$ Now you can hopefully see why I asked my original question.

However, now I can see that the small triangles are both similar to the large triangle, which implies that the small triangles are similar to each other, so their angle measures are the same. Then these triangles must be congruent because they also have a side of equal length. Hence, it must be that $$a=x$$ and $$b=y$$.

Thank you, however, to those of you who answered my original question.