Minimum distance from a point to a line

Text

Given the point $P(3,-2)$ and the line $y=2x+1$, find the point $R$ on the line at minimum distance from $P$, setting up the problem as a minimization.

Solution

A generic point on the line is $R=(x,\,2x+1)$. The squared distance from $P$ is $d^2(x)=(x-3)^2+(2x+1-(-2))^2=(x-3)^2+(2x+3)^2.$ Minimizing $d^2$ is equivalent to minimizing $d$. Differentiating: $\frac{d}{dx}\,d^2=2(x-3)+2(2x+3)\cdot2=2x-6+8x+12=10x+6.$ Setting to zero: $10x+6=0\Rightarrow x=-\dfrac35=-0.6$, and $y=2(-0.6)+1=-0.2$. Hence $R=\left(-\tfrac35,-\tfrac15\right)=(-0.6,\,-0.2).$ The second derivative $20>0$ confirms the minimum. The minimum distance: $d_{min}=\frac{|2\cdot3-(-2)+1|}{\sqrt{2^2+1^2}}=\frac{|6+2+1|}{\sqrt5}=\frac{9}{\sqrt5}\approx 4.025.$ $\boxed{\;R=(-0.6,\,-0.2),\qquad d_{min}=\dfrac{9}{\sqrt5}\approx 4.025\;}$