Answer:
(-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)
