Question

Find the point on the line
y = 9x closest to the point (1, 0).

Answer 1

We can solve this two ways, first using calculus, then without:

In both cases, first we simplify the equation of the line to 3x + y - 1 = 0
and then to y = 1 - 3x

First way, using calculus:

The distance from any point p on the line to (1,5)
is given by sqrt ( (x-1)^2 + (y-5)^2 )
We wish to minimize that.
We can ignore the sqrt, since for positive d, the minimum of d and minimum of d^2 occur for the same d.

We substitute the value of y from the equation of the line into the formula of the distance, take the derivative, and find where it is 0.

(x-1)^2 + (y-5)^2
(x-1)^2 + (1-3x-5)^2
x^2 - 2x + 1 is the square of the 1st term
the second term = -3x-4 which has the same square as 3x+4, which is 9x^2 +24x +16
so we have the sum of the two:
10x^2 + 22x + 17
Derivative is
20x + 22
Setting that to 0, we get 20x = -22 and x = -11/10
y = 1 - 3x = 1 + 33/10 or
So the point is (-1.1, 4.3)

Second way:

Another way to find it without using calculus,
is to derive the line through 1,5, perpendicular to the first line.

First line has slope -3, so new line will have slope 1/3
1/3 (1-x) = (5-y)
1/3 - x/3 = 5-y
1 - x = 15 - 3y
3y = x +14
Plugging in y from the first line, we get
3 (1 - 3x) = x + 14
3 - 9x = x +14
-10x = 11
x = -11/10
which agrees with the other solution.