Formula for a straight line y-y1=m(x-x1), where m is the gradient substituting in the values given to find the gradient we get 4-1=m(3+2), therefore m= 3/5the midpoint of the two points is ((x1 + x2)/2 , y1 + y2)/2)so the midpoint of AB is (0.5, 2.5)gradient of perpendicular bisector is the negative reciprocal of m which is -5/3using the equation of a line from above we get y-2.5=-5/3(x-0.5) Multiply both sides of the equation by 3 and we get 3y-7.5=-5(x-0.5)=-5x+2.5x Rearrange to get in the form stated in the question by adding 5x and adding 7.5 then we get 3y+5x = 10