The co-ordinates of the mid -point is [(8+5)/2 , (8+3)/2][8+52,8+32] or (13/2,11/2)(132,112) ; The slope m1 of the line passing through (5,3) and (8,8)(5,3)and(8,8) is (8-3)/(8-5)8−38−5 or5/353; We know the cond ition of perpendicularity of two lines is as m1*m2 = -1m1⋅m2=−1 where m1 and m2 are the slopes of the perpendicular lines. So the slope of the line will be (-1/(5/3))(−153) or -3/5−35 Now the equation of line passing through the mid point is (13/2,11/2)(132,112) is y-11/2 = -3/5(x-13/2)y−112=−35(x−132) or y=-3/5*x+39/10+11/2y=−35⋅x+3910+112 or y + 3/5*x = 47/5y+35⋅x=475 or 5*y+3*x=475⋅y+3⋅x=47[Answer]
