Jump to content
Rachel Maddiee

Point and a line

Recommended Posts

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 by mathematic
latex setup

Share this post


Link to post
Share on other sites

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.

Share this post


Link to post
Share on other sites

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.

Share this post


Link to post
Share on other sites

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).

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.