Expert: Paul Klarreich - 1/18/2010

Question
hello sir.

could you help me answering this problem.

Find the shortest distance from the origin to the line 3x + y = 6, and find the point P on the line closest to the origin. Then show that the origin lies on the line perpendicular to the given line at P.

thank you so much.

by the way, i already answered this question but i'm not sure. what i got is 3/5 square root of 10 units, and (9/5 , 3/5)

Answer
Questioner: samantha
Country: Philippines
Category: Calculus
Private: No
Subject: applications of absolute extrema
Question: hello sir.

could you help me answering this problem.

Find the shortest distance from the origin to the line 3x + y = 6, and find the point P on the line closest to the origin. Then show that the origin lies on the line perpendicular to the given line at P.

thank you so much.

by the way, i already answered this question but i'm not sure. what i got is 3/5 square root of 10 units, and (9/5 , 3/5)
.................................................
There is a standard formula:

www.intmath.com/Plane-analytic-geometry/Perpendicular-distance-point-line.php


So your distance should be
   | 3(0) + (0) - 6 |
d = ------------------ = 6/sqrt(10), which is your answer.
   sqrt(9 + 1)

Now the DEFINITION of d(pt,line) is the perpendicular from the point to the line.  So your questions are kind of backwards.  they should be:

The line 3x + y = 6 has slope = -3, so the perpendicular has slope +1/3.

Now if it p.t. (0,0), write:

y - y0 = m(x - x0), and use  m = 1/3, y0 = 0,  x0 = 0.

y = x/3.  or  3y = x  or - x + 3y = 0

Now solve the equations:

3x +  y = 6
- x + 3y = 0
--------------

3x +  y = 6
-3x + 9y = 0
--------------
    10y = 6
     y = 3/5

     x = 3y = 9/5

So you did OK.