# Find the point on the line 6x+2y-2=0 which is closest to the point (1,5)?

Find the point on the line 6x+2y-2=0 which is closest to the point (1,5)

Relevance

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.

• If we draw a line from the closest element on the function to (-a million,-4), the line could be commonly used to the tangent at that element -6x + 2y + 6 = 0 y = 3x - 3 y' = 3 the tangent has a slope of three, we prefer a line perpendicular to this that passes with the aid of (-a million,-4) y* = (-a million/3)x + b sub interior the ordered pair -4 = (-a million/3)(-a million) + b b = -13/3 y* = -a million/3x - 13/3