2. Given a directed line segment with endpoints A(3, 2) and B(6, 11), what is the point thatdivides AB two-thirds from A to B?

Answer:The point of division is (5 , 8)Step-by-step explanation:* Lets explain how to solve the problem- If point (x , y) divides the line whose endpoints are [tex](x_{1},y_{1})[/tex]  and [tex](x_{2},y_{2})[/tex] at ratio [tex]m_{1}:m_{2}[/tex] , then  [tex]x=\frac{x_{1}m_{2}+x_{2}m_{1}}{m_{1}+m_{2}}[/tex] and  [tex]y=\frac{y_{1}m_{2}+y_{2}m_{1}}{m_{1}+m_{2}}[/tex]* Lets solve the problem- The directed line segment with endpoints A (3 , 2) and B (6 , 11)- There is a point divides AB two-thirds from A to B∵ The coordinates of the endpoints of the directed line segments    are A = (3 , 2) and B = (6 , 11)∴ [tex](x_{1},y_{1})[/tex] is (3 , 2)∴ [tex](x_{2},y_{2})[/tex] is (6 , 11)∵ Point (x , y) divides AB two-thirds from A to B- That means the distance from A to the point (x , y) is 2/3 from   the distance of the line AB, and the distance from the point (x , y)   to point B is 1/3 from the distance of the line AB∴ [tex]m_{1}:m_{2}[/tex] = 2 : 1∵ [tex]x=\frac{(3)(1)+(6)(2)}{2+1}=\frac{3+12}{3}=\frac{15}{3}=5[/tex]∴ The x-coordinate of the point of division is 5∵ [tex]y=\frac{(2)(1)+(11)(2)}{2+1}=\frac{2+22}{3}=\frac{24}{3}=8[/tex]∴ The y-coordinate of the point of division is 8∴ The point of division is (5 , 8)