Find the point on the line 6x+y=9 that is closest to the point (3,8)

Question

How do I solve it?

Answer 1

Since 6x+y = 9, y = -6x + 9

 

So, a point, Q, on the line can be expressed as (x, -6x+9)

 

We want to minimize the distance frpm P = (3,8) to Q

 

                                   = √[(x-3)²+(-6x+1)²]

 

                                   = √[x²-6x+9+36x²-12x+1]

 

                                   = √[37x²-18x+10]

 

It's enough to minimize f(x) = 37x²-18x+10

 

   To minimize f(x), take the derivative and set it equal to zero.

 

     f'(x) = 74x - 18

 

     f'(x) = 0 when x = 9/37  (critical point)

 

     If x < 9/37, f'(x) < 0.  So, f(x) is decreasing when x < 9/37.

 

     If x > 9/37, f'(x) > 0.  So, f(x) is increasing when x > 9/37.

 

     Therefore, f(x) is minimized when x = 9/37.

 

The point Q on the line that is closest to (3,8) is:

 

    Q = (x, -6x+9) = (9/37, 279/37))

 

                 

Answer 2

Here is another possible way to find the point on the line y=-6x+9 that is closest to (3,8).

 

Let's say the x coordinate of the unknown point is a, then the point will have coordinates (a, -6a+9).

 

Now imagine a circle with center at (3,8) that has some radius R.

 

The equation of that circle would be (x-3)² + (y-8)² = R².

 

If we implicitly differentiate the equation of the circle and solve for y' we get y' = (3-x)/(y-8)  --notice that R is a constant so is not involved in the formula for y'.

 

That is a formula for slopes of tangent lines to the circle. When the given line is tangent to the circle, the point (a, -6a+9) will be the point of tangency--we have deemed that point to be closest to (3,8); the radius of the circle is perpendicular to point of tangency. Just let x=a and y=-6a+9 in the formula for y', setting y' = -6 (the slope of the line).

 

So, y' = (3-a)/(-6a+1) = -6

 

3-a = 36a - 6

 

37a = 9

 

a = 9/37

-6a+9 = 279/37

 

The point (9/37, 279/37) is on the line and on the circle. Note we have not actually determined the radius of the circle but it could be done.

 

Not sure this is any easier than finding the equation of the line perpendicular to the given line passing through (3,8) then finding the point of intersection. Just a different way to think about it and certainly involves calculus.

 

Answer 3

As Ismael said... the closest distance is a straight line.

You need to find the equation of the line perpendicular to

6x+y=9 going through (3,8) and find the intersection point.

 

6x+y=9

y = -6x +9

slope m=-6

 

perpendicular slope = -1/(-6) = 1/6

 

Using the point slope form y-y₁=m(x-x₁)

with m=1/6 and (x₁,y₁) = (3,8)

 

y-8 = (1/6)(x-3)

6y-48 = x-3

x-6y=-45    This is the perpendicular line through (3,8)

 

6x + y = 9

x - 6y = -45      solve by elimination

---------------

 

36x + 6y = 54     6 times equation 1

    x - 6y = -45

-----------------

37x = 9

 

x = 9/37

 

Substitute that back into either equation and solve for y

to get your intersection point

 

 

Answer 4

The closest point is the intersection of line 6x + y = 9 and the perpendicular line to it that passes through point (3,8).

When you have a line y = m1x + b with slope m1 then a perpendicular line to it has a slope m2 = -1/m1

The slope of line 6x + y = 9 is -6 therefore the slope of a perpendicular line to it is m = -1/-6

You now have the slope of the perpendicular line (-1/-6) passing through point (3,8).

You use the slope-point formula to get the formula for the perpendicular line.

Now you have two lines. Solve system of equations to find the intersection point (the solution to the problem).