Find the point $(x_0, y_0)$ on the line $ax + by = c$ that is closest to the origin. Find the point  $(x_0, y_0)$  on the line  $ax + by = c$  that is closest to the origin. 
According to this source, I thought that $\left( -\frac{ac}{a^{2}+b^{2}}, -\frac{bc}{a^{2}+b^{2}} \right)$ was the point but it doesn't seem to be correct. Thanks for any help.
 A: \begin{array}{cccc}
  ax+by &=& c & \cdots \cdots (1) \\
  bx-ay &=& 0 & \cdots \cdots (2)
\end{array}
$(1)$ is equation of the given line,
$(2)$ is the equation of the normal passing through the origin
$(1)\cap(2)$ is the required point.
$(1)\times a+(2)\times b$,
$$(a^{2}+b^{2})x=ac$$
$(1)\times b-(2)\times a$,
$$(a^{2}+b^{2})y=bc$$
$$\therefore \quad
(x,y)=\left( \frac{ac}{a^{2}+b^{2}}, \frac{bc}{a^{2}+b^{2}} \right)$$
A: $$(h-0)^2+\left(\dfrac{c-ah}b\right)^2=\dfrac{(a^2+b^2)h^2-2cah+c^2}{b^2}$$
Now $(a^2+b^2)h^2-2cah=\left(h\sqrt{a^2+b^2}-\dfrac{ca}{\sqrt{a^2+b^2}}\right)^2-\dfrac{(ca)^2}{a^2+b^2}\ge-\dfrac{(ca)^2}{a^2+b^2}$
the minimum value occurs iff
the equality occurs if $$h\sqrt{a^2+b^2}-\dfrac{ca}{\sqrt{a^2+b^2}}=0\iff h=?$$
Can you take it from here?
A: $$ax_0+by_0=c$$
Thus,
$$y_0=\frac{c-ax_0}{b}$$
The distance of $(x_0,y_0)$ from $(0,0)$ is, $d=\sqrt{x_0^2+y_0^2}$
Thus,
$$d=\sqrt{x_0^2+\left(\frac{c-ax_0}{b}\right)^2}=\frac{\sqrt{b^2x_0^2+c^2+a^2x_0^2-2acx_0}}{b}$$
To minimize $d$, taking the derivative with respect to $x_0$,
$$\frac{d}{dx_0}(d)=\frac{2(a^2+b^2)x_0-2ac}{2b\sqrt{b^2x_0^2+c^2+a^2x_0^2-2acx_0}}=0$$
Thus,
$$x_0=\frac{ac}{a^2+b^2}$$
Correspondingly,
$$y_0=\frac{bc}{a^2+b^2}$$
It can be verified that this is indeed the point of minimum. Intuitively, it is clear that it must be so as a point with minimum distance must exist while a point with maximum distance from the origin does not exist.
A: Calling $\bar v =(a,b), \bar x=(x,y)$, the line has equation $\bar v \cdot \bar x=c$.
Cauchy-Schwarz gives that $\vert \bar v\vert  \cdot \vert \bar x\vert \ge c$, with equality when $\bar v, \bar x$ are linearly dependent.
Setting $\bar x = \lambda\bar v$ and plugging it into $\bar v \cdot \bar x=c$, we see that equality is attained at $\bar x = \frac{c}{\vert \bar v \vert ^2}\bar v$, where $\vert \bar x \vert ^2=\frac{c^2}{\vert \bar v \vert ^2}$.
This corresponds to $(x,y)=\left(\frac{ac}{a^2+b^2},\frac{bc}{a^2+b^2}\right)$ and $d=\frac{c}{\sqrt{a^2+b^2}}$.
A: The requested point lies on the perpendicular ($(a,b)\to(b,-a)$) to the line drawn from the origin. Solve
$$\begin{align}ax+by&=c\\bx-ay&=0.\end{align}$$
A: Since given question has multivariable-calculus, I will use Lagrange multiplier. Define $f,g$ as
\begin{align}
f(x,y)&=x^2+y^2\\
g(x,y)&=ax+by-c=0,
\end{align}
Then there exists $\lambda\in \mathbb{R}$ such that
$$
(2x,2y)=\lambda(a,b)
$$
and so $2x=\lambda a$ and $2y=\lambda b$. Then
$$
ax+by-c=\frac{1}{2}a^2\lambda +\frac{1}{2}b^2\lambda -c=0
$$
and so
$$
\lambda=\frac{2c}{a^2+b^2}
$$
It means that $f(x,y)$ has a critical point at $(x,y)=\left(\frac{ac}{a^2+b^2},\frac{bc}{a^2+b^2}\right)$. It is the closest point.
