For a linear relationship, y=mx+b, m=(y2-y1)/(x2-x1) is a constant, all lines have a constant velocity, called the slope, m.
(4-3)/(4-2)=(5-4)/(6-4)=1/2=m so
y=x/2+b, using any point we can solve for "b", the y-intercept (the value of y when x=0), I'll use point (2,3)
3=2/2+b
3=1+b
2=b so the line is:
y=x/2+2 or as it is expressed in your choices above:
Y=(1/2)x+2