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| 1. |
Trisection of the line segment joining (4,-1) and (-2,-3) |
| Answer» Let P and Q be the points of trisection as shown belowThen, AP : PB = 1 : 2and AQ : QB = 2 : 1\t\tHere,\xa0{tex}\\frac{m_{1}}{m_{2}}=\\frac{1}{2},{/tex}\xa0A(x1, y1) = (2, -3) and B(x2, y2) = (4, -1)\tFor internally,\xa0{tex}P=\\left(\\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\\right){/tex}\t{tex}\\Rightarrow \\quad P=\\left(\\frac{1 \\times 4+2 \\times 2}{1+2}, \\frac{1 \\times(-1)+2 \\times(-3)}{1+2}\\right){/tex}\t{tex}\\Rightarrow \\quad P=\\left(\\frac{4+4}{3}, \\frac{-1-6}{3}\\right) \\Rightarrow P=\\left(\\frac{8}{3}, \\frac{-7}{3}\\right){/tex}\tHere,\xa0{tex}\\frac{m_{1}}{m_{2}}=\\frac{2}{1}{/tex}, A(x1, y1) = (2, -3)\tand B(x2, y2) = (4, -1)\tFor internally,\xa0{tex}Q=\\left(\\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\\right){/tex}\t\t{tex}\\Rightarrow \\quad Q=\\left(\\frac{2 \\times 4+1 \\times 2}{2+1}, \\frac{2 \\times(-1)+1 \\times(-3)}{2+1}\\right){/tex}\t{tex}\\Rightarrow \\quad Q=\\left(\\frac{8+2}{3}, \\frac{-2-3}{3}\\right) \\Rightarrow Q=\\left(\\frac{10}{3}, \\frac{-5}{3}\\right){/tex} | |