Can someone please help me with this question?

9.\( d^2=(x-6)^2+(y-2)^2\), where \(y=6x+3\).   Or \(d^2=(x-6)^2+(6x+1)^2.\)  To minimize calculate derivative and set to 0.   \(2(x-6)+12(6x+1)=0\) leading to x=0 and y=3  or \(d^2=37\).

 

10.  Statement is incomplete.

Edited December 16, 20196 yr by mathematic
latex setup

Can you help me with the justification/explanations?

First equation is distance between given point and any point on any line.  Next replace y by function of x for given line.  Use calculus to get minimum distance and get particular point on line that is minimum.

Can you show me using this method?

The problem you have is different from the problem in the post (2.10) - distance between parallel lines.

Yes, it’s an example of the accepted method.

It looks to me the first step for two parallel lines is pick a point on one line and then find the distance to the other line.  An alternative (without calculus) to the solution I had described is to set up the general form of a line perpendicular to the given line and then adjust one parameter to make it pass through the given point.

Can you show me?

Can you show us what you've tried so far?

The shortest distance between a point and a line (which is what is meant by "the" distance between them) is along a line perpendicular to the line. The line is given by y= 6x+ 3 so has slope 6. A line perpendicular to that has slope -1/6. The line through (6, 2) with slope -1/6 is y= (-1/6)(x- 6)+ 2 so 6y= -x+ 6+ 12 or x+ 6y= 18

Substitute (two lines which intersect) y=6x+3 into x+6y=18 and you get:
x + 6(6x+3) = 18.
x + 36x +18 = 18
37x +18 =18
37x = 0
x = 0
So that means that x=0 at the point where the lines y=6x+3 into x+6y=18 intersect.
Substitute y = 6x + 3 
y = 6(0) + 3 = 3
y = 3
The point of intersection is  (0, 3).

Then, use the distance formula to find the exact distance between (0,3) and 6,2)
d^2 = (x2 - x1)^2 + (y2 - y1)^2
So the distance between (0, 3) and (6, 2), squared, D^2= (0- 6)^2+ (3- 2)^2= 37 and then D= square root (37).

Hooray!!!!