Question

Find the point P on the line yequals=33x that is closest to the point (60 comma 0 )(60,0). What is the least distance between P and (60 comma 0 )(60,0)​?

Answer 1

Answer:

$18\sqrt{10} units$

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

$=\sqrt{(60-a)^2+(0-3a)^2}\\D =\sqrt{10a^2-120a+3600}$

To minimize D, we find its derivative

$\dfrac{dD}{da}=\dfrac{10a-60}{\sqrt{10a^2-120a+3600} }\\ Setting \dfrac{dD}{da}=0\\10a-60=0\\10a=60\\a=6$

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).

$D =\sqrt{10(6)^2-120(6)+3600}\\=\sqrt{3240}\\=18\sqrt{10} units$