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)

h/o/e/s mad

step-by-step explanation:

si si si

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.