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

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SUMMARY

The smallest distance from point P(-8, 4) to the line y = 6x is calculated using the formula for distance from a point to a line, yielding a result of approximately 8.54. The parameters used in the calculation are A = 6, B = -1, m = -8, and n = 4. The discussion also explored an alternative method using vector analysis, although the setup was not completed. The squared distance formula D² = (x + 8)² + (6x - 4)² was derived, leading to the critical value x = 16/37.

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karush
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$\tiny{231.12.3.63}$
$\textsf{ Determine the smallest distance between point P and the line L through the origin}\\$
$\textsf{$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 ?}\\$
$\textit{also, was going to try to solve this with a vector but couldn't seem to to set it up...}$
 
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The square of the distance is:

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

Critical value is from:

$$74x-32=0\implies x=\frac{16}{37}$$

$$D_{\min}=\frac{52}{\sqrt{37}}$$

Just as you found. :D
 
thanks... wouldn't have thot of that

nice thing about MHB...:cool:
 

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