Yahoo Answers: Answers and Comments for Find the point on the line 2x+3y+4 =0 which is closest to the point ( 4, 5 )? [Mathematics]
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From Anonymous
enUS
Tue, 06 Dec 2011 11:11:34 +0000
3
Yahoo Answers: Answers and Comments for Find the point on the line 2x+3y+4 =0 which is closest to the point ( 4, 5 )? [Mathematics]
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https://answers.yahoo.com/question/index?qid=20111206111134AA0fDfT
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From larrauri: locate the component on the line 2x + 3y  a m...
https://answers.yahoo.com/question/index?qid=20111206111134AA0fDfT
https://answers.yahoo.com/question/index?qid=20111206111134AA0fDfT
Sat, 19 Nov 2016 04:44:01 +0000
locate the component on the line 2x + 3y  a million = 0, .................. [a million] that's closest to the component (a million, a million)? xintercept, x= a million/2, yintercept, y = a million/3 for that reason the component closest to [a million] is, (0, a million/3). answer.

From Keith Roberts: The shortest distance between a point and a li...
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https://answers.yahoo.com/question/index?qid=20111206111134AA0fDfT
Tue, 06 Dec 2011 11:52:45 +0000
The shortest distance between a point and a line is the distance that is coincident with the line passing through the point and perpendicular to the line. So in this problem, our strategy will be to (1) find the equation of the perpendicular line passing through (4,5), and then (2) find the point of intersection between it and the given line.
(1) To find the equation of the line perpendicular to L: 2x+3y+4=0, passing through (4, 5), we note we already have a point on that line, namely (4,5), and so only need a slope of that line in order to find the equation. Using L's slope, we can find the slope of the line perpendicular by recognizing that perpendicular lines slopes are negative reciprocals. Thus, rewriting L in slopeintercept form, we obtain y = (2/3)x  (4/3), and the slope of L is 2/3. Thus the slope of the line perpendicular to L is 3/2.
Now, to find the equation of the line perpendicular to L passing through the point (4,5), we use that point and the slope of the line, 3/2, to find that the equation of the line perpendicular to L is y = (3/2) x + 1.
(2) Finally, we can find the the point of intersection of L and the line perpendicular. This can be found using substitution (or any other method of solving linear systems of two equations in two unknowns) to find x = 14/13. Thus, y = 8/13 and the point on L closest to (4,5) is (14/13,8/13).
To help reinforce your understanding of these concepts, I've searched and found a webpage and a video tutorial that address problems similar to this one, and I thought they might be helpful to you. I've listed them below.
As always, if you need more help, please clarify where you are in the process and what's giving you trouble. I'd be more than happy to continue to assist.

From Moise Gunen: Any point to line 2x+3y+4 =0 has coordinates...
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Tue, 06 Dec 2011 11:25:05 +0000
Any point to line 2x+3y+4 =0 has coordinates:
(x, (2x4)/3 )
Let be a mobile point M (x, (2x4)/3 ) to this line
Distance D between point (4, 5) and M is:
D^2 = (x4)^2 + ((2x4)/3 + 5)^2 =
x^2  8x + 16 + (2x+ 11)^2 / 9 =
(1+4/9)x^2  x(8  44/9) + 16 + 121/9 =
(13/9)x^2  (28/9)x + 265/9
For x = (28/9)/(2*13/9) = (28/9)*(9/26) = 14/13 you have a local minimum for this parabola
For x = 14/13 you get y = (2*(14/13)4)/3 = (28/13 52/13)/3 = 8/13
Closest point is
(x,y) = (14/13, 8/13)