Point and a line

[Rachel Maddiee]

Can someone please help me with this question?

[mathematic]

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, 2019 by mathematic
latex setup

[Rachel Maddiee]

Can you help me with the justification/explanations?

[mathematic]

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.

[Rachel Maddiee]

Can you show me using this method?

[mathematic]

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

[Rachel Maddiee]

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

[mathematic]

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.

[Rachel Maddiee]

Can you show me?

[zapatos]

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

[Rachel Maddiee]

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).

[mathematic]

Hooray!!!!