Smallest Distance $P(-8,4)$ to Line $y=6x$ at Origin

[karush]

$\tiny{231.12.3.63}$
$\textsf{ Determine the smallest distance between point P and the line L through the origin}\\$
$\textsf{$P(-8,4)$ L is $y=6x$ }$ \\
$\textsf{$\therefore A=6, B=-1, m=-8, n=4$}$
\begin{align*}\displaystyle
d&=\frac{|Am+Bn+C|}{\sqrt{A^2+B^2}}\\
&=\frac{|(6)(-8)+(-1)(4)|}{\sqrt{36+1}} \\
&=\frac{52}{\sqrt{37}}\approx8.54
\end{align*}
$\textit{Demos indicated that the answwer to this was $\approx8.43 $ so ?}\\$
$\textit{also, was going to try to solve this with a vector but couldn't seem to to set it up...}$

 

[MarkFL]

The square of the distance is:

\(\displaystyle D^2=(x+8)^2+(6x-4)^2=37x^2-32x+80\)

Critical value is from:

\(\displaystyle 74x-32=0\implies x=\frac{16}{37}\)

\(\displaystyle D_{\min}=\frac{52}{\sqrt{37}}\)

Just as you found. :D

 

[karush]

thanks... wouldn't have thot of that

nice thing about MHB...