Minimum distance from a point to a line (2)

Text

Given the point $P(4,4)$ and the line $y=-2x+3$, 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 $R=(x,\,-2x+3)$. The squared distance from $P$ is $d^2(x)=(x-4)^2+(-2x+3-4)^2=(x-4)^2+(-2x-1)^2.$ Differentiating: $\frac{d}{dx}\,d^2=2(x-4)+2(-2x-1)(-2)=2x-8+8x+4=10x-4.$ Setting to zero: $10x-4=0\Rightarrow x=\dfrac25=0.4$, and $y=-2(0.4)+3=2.2$. Hence $R=\left(\tfrac25,\tfrac{11}5\right)=(0.4,\,2.2).$ The second derivative $20>0$ confirms the minimum. The minimum distance: $d_{min}=\frac{|-2\cdot4-4+3|}{\sqrt{(-2)^2+1^2}}=\frac{|-8-4+3|}{\sqrt5}=\frac{9}{\sqrt5}\approx 4.025.$ $\boxed{\;R=(0.4,\,2.2),\qquad d_{min}=\dfrac{9}{\sqrt5}\approx 4.025\;}$