Biet rang A,B la 2 diem thuoc do thi ham so y=-4/3x+4 
a,tim tung do cua diem A biet hoanh do =0
xA = 0 => yA = -4/3. 0 + 4 = 4
=> A(0; 4)
b,tim hoanh do cua diem B biet tung do = 0
=> yB = 0
=> 0 = -4/3.xB + 4
=> xB = 3
=> B(3; 0)
c,tinh khoang cach AB theo do dai don vi
AB = √[(0 - 3)² + (4 - 0)²] = 5