Find the point on the line 6x + 3y-3 =0 which is closest to the point (3,1).

Note: Your answer should be a point in the xy-plane, and as such will be of the form (x-coordinate,y-coordinate)

To find the point on the line 6x + 3y - 3 = 0 that is closest to the point (3, 1), we can use the concept of perpendicular distance.

First, let's rearrange the given equation to the slope-intercept form (y = mx + c):

6x + 3y - 3 = 0
3y = -6x + 3
y = -2x + 1

From the equation, we can see that the slope of the line is -2 (the coefficient of x).

To find the point on the line that is closest to (3, 1), we draw a perpendicular line from (3, 1) to the given line. The point of intersection of these two lines will be the required point.

Let's start by finding the equation of the perpendicular line. Since the perpendicular line has a slope that is the negative reciprocal of -2, the slope of the perpendicular line will be 1/2.

Using the point-slope form (y - y1 = m(x - x1)), we can write the equation of the perpendicular line passing through point (3, 1):

y - 1 = (1/2)(x - 3)

Rearranging this equation, we get:

y = (1/2)x - 3/2 + 1
y = (1/2)x - 1/2

Now, we have a system of equations:

y = -2x + 1 (equation of the given line)
y = (1/2)x - 1/2 (equation of the perpendicular line)

To find the point of intersection, we solve this system of equations. By setting the two y-values equal, we can solve for x:

-2x + 1 = (1/2)x - 1/2

Simplifying the equation:

-2x - (1/2)x = -1/2 - 1

-4x - x/2 = -2/2 - 1/2

(-8x - x) / 2 = -3/2

-9x / 2 = -3/2

Solving for x:

-9x = -3
x = -3 / -9
x = 1/3

Now, substitute the value of x back into one of the equations to find the y-coordinate. Let's use the equation of the given line:

y = -2(1/3) + 1
y = -2/3 + 1
y = -2/3 + 3/3
y = 1/3

Therefore, the point on the line 6x + 3y - 3 = 0 closest to the point (3, 1) is (1/3, 1/3).

what a stupid note. Of course it will be a point with (x,y) coordinates! This is a calculus class - you know all that already.

Given a point (x,y) on the line, y=1-2x

The distance from (x,1-2x) to (3,1) is

d^2 = (x-3)^2 + (1-(1-2x))^2 = 5x^2-6x+9

so, we want minimum d, when dd/dx = 0:

2d dd/dx = 10x-6
dd/dx = 10x-6/2sqrt(blah blah)

dd/dx=0 when x = 3/5.

So, minimum d^2 is

5(3/5)^2 - 6(3/5) + 9 = 36/5
minimum d is 6/√5

Or, as we all know, the distance from a point (x,y) to a line ax+by+c=0 is

|ax+by+c|/√(a^2+b^2) = (6(3)+3(1)-3)/√(36+9) = 18/√45 = 6/√5