SOLUTION: Find the point on the line y=5x+1 that is closest to the point (3,5) .

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Question 1160834: Find the point on the line y=5x+1 that is closest to the point (3,5) .
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!

Find the perpendicular line through (3, 5) that intersects the line y=5x+1
the perpendicular line has slope -(1/5), the negative reciprocal.
so y-y1=m(x-x1) point slope formula, m=slope and (x1, y1) the point.
y-5=(-1/5)(x-3)
y=(-1/5)x+3/5+5
y=(-1/5)x+(28/5)
Those two lines intersect at a point when (-1/5)x+(28/5)=5x+1
or (26/5)x=(23/5)
or x=(23/26)
when x=(23/26), y=141/26 using the 5x+1
and x=(23/26), y=-23/130+(728/130), or (705/130), which is 141/26
One could use the distance formula, but the perpendicular line to the intersection of the two will yield the closest point.
((-23/26), (141/26))
Answer by ikleyn(52790)   (Show Source): You can put this solution on YOUR website!
.

            I will present here another way to solve the problem. 


On the line (x,5x+1) we want to find the point closest to (3,5).


Write the distance formula (the square of distance formula)


    d^2 = (x-3)^2 + (5x+1-5)^2 = 

          (x-3)^2 + (5x-4)^2 = x^2 - 6x + 9 + 25x^2 - 40x + 16 = 

                             = 26x^2 - 46x + 25.


The minimum of this quadratic form is at  x =  =  = .


ANSWER.  The closest point is  ( , ) = ( , ).

Solved.



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