3 points lie in the same straight line. Then why is this formula valid?

Question

I am given that the points $(0,1)$, $(x,y)$ and $(\phi(x,y),0)$ are all lying in the same straight line from the north pole of the circle $S^1=\{ (x,y)\in \mathbb{R}: x^2+y^2=1 \}$ to the point $(\phi(x,y),0)$ which is a point on the $y=0$ axis. The middle point $(x,y)$ lies on the circle. Then why is it true that $$\frac{x}{\phi(x,y)}+y=1$$

Answer 1

If you know about determinants, then one nice trick to know is that three points $(a_1,b_1),(a_2,b_2)$ and $(a_3,b_3)$ are collinear if and only if $$\det\begin{bmatrix} a_1 & b_1 & 1\\a_2 & b_2 & 1\\ a_3 & b_3 & 1\end{bmatrix}=0.$$

So given the information in your question, you know that $$\det\begin{bmatrix} 0 & 1 & 1\\x & y & 1\\\phi(x,y) & 0 & 1\end{bmatrix}=0,$$ which says that $\phi(x,y)-y\phi(x,y)-x=0.$ Rearranging (and assuming $\phi(x,y)\neq 0$), gives $\frac{x}{\phi(x,y)}+y=1$.

Answer 2

For any fixed $a,b,c$ (with $(a,b)\ne(0,0)$) the equation $ax+by=c$ describes a line. If we let $a=\frac1{\phi(x,y)}$ (assuming $\phi(x,y)\ne0$, of course) and $b=1$ and $c=1$, we see that $(0,1)$ is on this line and $(\phi(x,y),0)$ is on this line. Since there is only one line through two distinct points, the point $(x,y)$ must also fulfill the equation of this line.

Answer 3

Consider the line from $(0,1)$ to $(x,y)$ this has the slope $\frac{dy}{dx}=\frac{1-y}{x}$.

Thus $\phi(x,y)$ is $$\Delta x = \frac{dx}{dy}\Delta y$$ away from x, where $\Delta y$ is here the difference from in y coordinate of the point $(x,y)$ and $(\phi(x,y),0)$ and thus $\Delta y = y$

Therefore $$\phi(x,y)=x+\frac{dx}{dy}\cdot y = x+\frac{x}{1-y}\cdot y=x+\frac{xy}{1-y}$$ Devide y $x$: $$\frac{\phi(x,y)}{x}=1-\frac{y}{1-y}=\frac{1}{1-y}$$ The inverse: $$\frac{x}{\phi(x,y)}=1-y$$ and thus $$\frac{x}{\phi(x,y)}+y=1$$

Answer 4

I think you can prove it with similar triangles.

Let: $A:(0,1)$, $B:(x,y)$, $C:(\phi,0)$, $D:(x,0)$, and $O: (0,0)$, then We have: $$\overline{CD} : \overline{CO} =\overline{BD} : \overline{AO}$$ $${{\phi - x} \over {\phi}} ={{y} \over {1}}$$ $${x \over \phi}+y=1 $$

Answer 5

Let the points $(0,1)$, $(x,y)$, and $(\phi,0)$ be labeled $A$, $B$, $C$. Then the segments $AC$ and $AB$ have the same slope, since the three points lie on the same line. Calculate rise-over-run for each segment to assert: $$ {1-0\over 0-\phi} = \mbox{slope} ( AC)= \mbox{slope}(AB) = {1 - y \over 0 - x} $$ and simplify. The fact that $B$ lies on a circle isn't used.

Answer 6

We can naturally describe the situation using the parameterization of a line. In general, the parameterization of a line is defined by a point $\vec{c}_0$ and a direction $\vec{v}$, i.e. $$ \vec{c}(t) = \vec{c}_0 + t\vec{v}, $$ where $t$ ranges from $-\infty$ to $\infty$.

In this case, we can describe the line as starting from $(0,1)$ and going in the direction from $(0,1)$ to $(x,y)$: $$ \vec{c}(t) = \begin{bmatrix} 0 \\ 1\end{bmatrix} + t\left( \begin{bmatrix} 0 \\ 1\end{bmatrix} - \begin{bmatrix} x \\ y\end{bmatrix}\right) = \begin{bmatrix} -tx \\ 1+t(1-y)\end{bmatrix}. $$

Now we can set the left-hand side to $(\phi(x,y),0)$: $$ \begin{bmatrix} \phi(x,y) \\ 0\end{bmatrix} = \begin{bmatrix} -tx \\ 1+t(1-y)\end{bmatrix}, $$ which gives you a system of two equations $$ \phi(x,y) = -tx $$ and $$ 0=1+t(1-y). $$ From there, you can eliminate $t$ for the desired expression. The benefit of using this method is that it generalizes to three dimensions, for example.