Question 416991
Solution; First find the equation of line who pases on this two points;
(y+3)/6+3=(x+4)/4+4; (y+3)/6=(x+4)/8; 4(y+3)=3(x+4); 3x-4y=0; y=3x/4.
The midpoint is;x=(-4+4)/2=0; y=(-3+6)/2=1.5 ; (0,1.5) The equation of the line equidistant is;  y-1.5=-4/3(x-0); y=-4x/3+1.5 or 8x+6y-9=0.