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)
- MathMan TGLv 71 decade agoBest Answer
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
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)
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.
- okamotoLv 43 years ago
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