Answer:
![18√(10)$ units](https://img.qammunity.org/2021/formulas/mathematics/college/1avht1g8clor7qqhkr8n5fhkk01b46glab.png)
Explanation:
We are given the equation of the line y=3x and a point, say Q(60,0) outside of that line.
We want to find the point on the line y=3x which is closest to Q.
Let P(x,y) be the desired point. Since it is on the line y=3x, it must satisfy the line.
If x=a, y=3a, so the point P has the coordinates (a,3a).
Distance between point Q and P
![=√((60-a)^2+(0-3a)^2)\\D =√(10a^2-120a+3600)](https://img.qammunity.org/2021/formulas/mathematics/college/we3qiqy5otavlhu1f9izogufpi6mp78lxp.png)
To minimize D, we find its derivative
![(dD)/(da)=(10a-60)/(√(10a^2-120a+3600) )\\$Setting (dD)/(da)=0\\10a-60=0\\10a=60\\a=6](https://img.qammunity.org/2021/formulas/mathematics/college/k2ol90h58cqk7te37kgu2sjhehx9pkg5eb.png)
Therefore, the y-coordinate for P is 3*6=18.
The point P=(6,18).
Next, we calculate the distance between P(6,18) and (60,0).
![D =√(10(6)^2-120(6)+3600)\\=√(3240)\\=18√(10)$ units](https://img.qammunity.org/2021/formulas/mathematics/college/aapem5l6u2i1fp4apaiwj75a7ycu4i3xe1.png)