SOLUTION: Find the point on the line y=4x that is closest to the point P=(1,2).

Click here to see ALL problems on Points-lines-and-rays

Question 1160833: Find the point on the line y=4x that is closest to the point P=(1,2).
Found 3 solutions by Boreal, MathLover1, ikleyn:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website!
The line perpendicular to y=4x will contain that point
that perpendicular line is y=-1/4 * x, the negative reciprocal slope
it goes through (1, 2)
point slope formula y-y1=m(x-x1), m=slope, (x1, y1)= point.
y-2=(-1/4)(x-1)
y=(-1/4x)+9/4
find where the two lines intersect
4x=-(1/4)x+9/4
(17/4)x=9/4
x=9/17
first equation y=36/17 when x=9/17
second equation, y=-9/68+153/68=144/68=36/17, if divide top and bottom by 4
((9/17), (36/17))

Answer by MathLover1(20850)   (Show Source):
You can put this solution on YOUR website!
the point on the line that is closest to the point =(,) will be intersection point of the line and the line perpendicular to that passes through the point =(},)
the line have a slope
and the perpendicular line will have a slope negative reciprocal which is
then, the equation of the perpendicular line will be
......use given point =(},) to calculate

the equation of the perpendicular line is:

now find intersection point solving the system:
.........eq.1
....eq.2
-----------------------------------------

....both sides multiply by

-> coordinate
-> coordinate
intersection point: =(,)

Answer by ikleyn(52803)   (Show Source):
You can put this solution on YOUR website!
.

            I will present here another way to solve the problem.

On the line (x,4x) we want to find the point closest to (1,2). Write the distance formula (the square of distance formula) d^2 = (x-1)^2 + (4x-2)^2 = = x^2 - 2x + 1 + 16x^2 - 16x + 4 = = 17x^2 - 18x + 5. The minimum of this quadratic form is at x = = = . ANSWER. The closest point is ( , ) = ( , ).

Solved.