# Which point is closest to a specific line?

• December 3rd 2009, 10:45 AM
s3a
Which point is closest to a specific line?
Question:
Find the point on the line http://gauss.vaniercollege.qc.ca/web...249f9be551.png which is closest to the point http://gauss.vaniercollege.qc.ca/web...3284e71581.png.

(I am not sure if this is calculus but my course is called calculus so if it isn't please don't ban me or anything because I am not sure if you need knowledge of calculus for this or not.) I have no idea how to do this. My two attemps were using the distance formula and making d = 0 which does not work and my other attempt was plugging in either x = 2 or y = -3 and getting the other respective coordinate and that didn't work. Someone please show me what to do.

Any help would be greatly appreciated!
• December 3rd 2009, 11:24 AM
Quote:

Originally Posted by s3a
Question:
Find the point on the line http://gauss.vaniercollege.qc.ca/web...249f9be551.png which is closest to the point http://gauss.vaniercollege.qc.ca/web...3284e71581.png.

(I am not sure if this is calculus but my course is called calculus so if it isn't please don't ban me or anything because I am not sure if you need knowledge of calculus for this or not.) I have no idea how to do this. My two attemps were using the distance formula and making d = 0 which does not work and my other attempt was plugging in either x = 2 or y = -3 and getting the other respective coordinate and that didn't work. Someone please show me what to do.

Any help would be greatly appreciated!
$d=\sqrt{(x-2)^2+\left(\frac{1}{2}x-\frac{3}{2}+3\right)^2}$
Remember to find the extema (max or min) you have to set the derivative to zero. So differentiate $d$ and find the minimum of $d$. Setting the distance formula to zero would imply that the given point is actually on the line.
I would do this a somewhat different way- the shortest distance from a point to a line is along the perpendicular to the line. To see that, think about constructing a right triangle with that perpendicular and any other point on the line. The perpendicular is a leg and the line to the other point is the hypotenuse=- and by the Pythagorean formula, $c^2= a^2+ b^2$ the hypotenuse is always longer than either leg.