Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 18

Determine the point on the line $6x + y = 9$ that is closes to the point $(-3,1)$
By using formula to the point $(-3,1)$ and $(x,y)$ from the line that is $(x,-6x+9)$

$\begin{equation} \begin{aligned} d &= \sqrt{(x+3)^2 + (-6x+9-1)^2} = \sqrt{(x+3)^2 + (-6x + 8)^2}\\ \\ d &= \sqrt{x^2 + 6x + 9 + 36x^2 - 96x + 64}\\ \\ d &= \sqrt{37x^2 - 90x + 73} \end{aligned} \end{equation}$


If we take the derivative of the distance, by using Chain Rule...


$\begin{equation} \begin{aligned} d' &= \frac{1}{2} (37x^2 - 90x + 73)^{-\frac{1}{2}} (74x - 90)\\ \\ d' &= \frac{37x - 45}{\sqrt{37x^2 - 90x + 73}} \end{aligned} \end{equation}$


when $d'= 0$,
$0 = 37x - 45$

Then, the critical number is $\displaystyle x = \frac{45}{37}$

when $\displaystyle x = \frac{45}{37}$, then
$\displaystyle y = - 6 \left( \frac{45}{37} \right) + 9 = \frac{63}{37}$
Therefore, the point closes to $(-3,1)$ on the line $6x + y = 9$ is $\displaystyle \left( \frac{45}{37}, \frac{63}{37} \right)$