Find the point on the line 2x+6y+1=0 which is closest to the point (-5,-3)

(-1.875, -1.625)

Step-by-step explanation:

We want to find the point on the equation 2*x + 6*y + 1 = 0 which is closest to the point (-5, -3)

Remember that the distance between two points (a, b) and (c, d) is:

D = √( (a - c)^2 + (b - d)^2)

First we can write our equation as a line in the slope intercept form:

y = -(2/6)*x - 1

Now we could minimize the equation:

D = √( (x - (-5))^2 + (y - (-3))^2) = √( (x + 5)^2 + (y + 3)^2)

D = √(  ( x + 5)^2 + (  -(2/6)*x - 1 + 3) ^2)

Then we can minimize:

D^2 = (x + 5)^2 + ( -(2/6)*x + 2)^2

       = x^2 + 10*x + 25  + (4/36)*x^2  - (8/6)*x + 4

       =  (1 + 4/36)*x^2 + (10 - 8/6)*x + 29

        = (40/36)*x^2 + (52/6)*x + 29

This is a quadratic equation whose leading coefficient is positive, then the minimum of this equation is at the vertex.

The vertex is at the x-value such that (D^2)' = 0

Then:

(D^2)' = 2*(40/36)*x + (52/6)

This needs to be zero then:

0 = 2*(40/36)*x + (52/6)

-52/6 = (80/36)*x

(-52/6)*(36/80) = x

-1.875 = x

And the y-value when x = -1.875 is:

y = -(2/6)*-1.875 - 1

y = -1.625

Then the point on the line 2x+6y+1=0 that is closest to (-5,-3) is the point (-1.875, -1.625)

hope this is what you were asking for!

step-by-step explanation: