Question 974087
(-3,2) and (3t,2t)

need to minimize the squared deviations of (2t-2)^2 and (3t+3)^2 

Take the first derivative and set it equal to zero.

2(2t-2)*2 + 2(3t+3)*3=0

8t-16 + 18t +18=0

26t+2=0
t=(-1/13)

(-3,2), (-3/13,-2/13)


Distance is sqrt [(36/13)^2) + (28/13)^2 ]=sqrt (7.67+ 4.64)=3.51 units

Try 0 for t
(-3,2) and (0,0)  ;;; distance sqrt (13)=3.61

The graph shows the line that the two points are on.  The curve is the distance between the two points for various values of t.  It is a minimum at (-2/13) for t.  The distance from that minimum to the line is minimized.

{{{graph(300,300,-5,5,-2,40,13x^2+4x+13,(2/3)x)}}}