Question

find the points of trisection of the line joining (2,-3) (4,5)​

Answer 1

The points of trisection are (8/3, -1/3) and (10/3, 7/3)  

• Let A = (2, -3), B = (4, 5)
• Let P divide AB in the ratio 1:2
• Coordinates of P are given by

                  $P=(\frac{mx_{2}+nx_{1}}{m+n},\frac{my_{2}+ny_{1}}{m+n})$

                  $P=(\frac{1(4)+2(2)}{1+2},\frac{1(5)+2(-3)}{1+2})$

                  $P=(\frac{8}{3},\frac{-1}{3})$

• Let Q divide AB in the ratio 2:1
• Coordinates of Q are given by

                  $Q=(\frac{mx_{2}+nx_{1}}{m+n},\frac{my_{2}+ny_{1}}{m+n})$

                  $Q=(\frac{2(4)+1(2)}{2+1},\frac{2(5)+1(-3)}{2+1})$

                  $Q=(\frac{10}{3},\frac{7}{3})$