Answer:
[tex]18\sqrt{10}$ units[/tex]
Step-by-step 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
[tex]=\sqrt{(60-a)^2+(0-3a)^2}\\D =\sqrt{10a^2-120a+3600}[/tex]
To minimize D, we find its derivative
[tex]\dfrac{dD}{da}=\dfrac{10a-60}{\sqrt{10a^2-120a+3600} }\\$Setting \dfrac{dD}{da}=0\\10a-60=0\\10a=60\\a=6[/tex]
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).
[tex]D =\sqrt{10(6)^2-120(6)+3600}\\=\sqrt{3240}\\=18\sqrt{10}$ units[/tex]
