# Math Help - Which point is closest to a specific line?

1. ## Which point is closest to a specific line?

Question:
Find the point on the line which is closest to the point .

(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!
Thanks in advance!

2. Originally Posted by s3a
Question:
Find the point on the line which is closest to the point .

(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!
Thanks in advance!
This is like an optimization problem. Presumably your familiar with finding the extrema of functions. In this case, you just want to find the minimum of the distance function. You're on the right track with the distance formula. Set the DERIVATIVE to zero.

$d=\sqrt{(x-2)^2+\left(\frac{1}{2}x-\frac{3}{2}+3\right)^2}$

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.

3. 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.

Here your line is -2x+ 4y+ 3= 0 which is the same as 4y= 2x- 3 or y= (1/2)x- 3/2: its slope is 1/2. The perpendicular line has slope -1/(1/2)= -2. Find the equation of the line through (2,-3) with slope -2 and then determine where it crosses -2x+ 4y+ 3= 0.