Find the coordinate of point which divide linesegment to (-2., 2)(2, 8

1.	Find the coordinate of point which divide linesegment to (-2., 2)(2, 8)
Answer» Let P (x1, y1) Q(x2, y2) and R(x3, y3) be the points which divide the line segment AB into four equal parts.Then, P divides AB in the ratio 1 : 3 internally.{tex}x=\\frac{mx_2+nx_1}{m+n}{/tex}\xa0{tex}\\therefore x _ { 1 } = \\frac { ( 1 ) ( 2 ) + ( 3 ) ( - 2 ) } { 1 + 3 }{/tex}{tex}= \\frac { 2 - 6 } { 4 } = - \\frac { 4 } { 4 } = - 1{/tex}{tex}y=\\frac{my_2+ny_1}{m+n}{/tex}{tex}y _ { 1 } = \\frac { ( 1 ) ( 8 ) + ( 3 ) ( 2 ) } { 1 + 3 }{/tex}{tex}= \\frac { 8 + 6 } { 4 } = \\frac { 14 } { 4 } = \\frac { 7 } { 2 }{/tex}So,\xa0{tex}\\mathrm { P } \\rightarrow \\left( - 1 , \\frac { 7 } { 2 } \\right){/tex}Also, Q divides AB in the ratio 1 : 1 i.e.Q is the mid point of AB{tex}x _ { 2 } = \\frac { - 2 + 2 } { 2 } = 0{/tex}{tex}y _ { 2 } = \\frac { 2 + 8 } { 2 } = \\frac { 10 } { 2 } = 5{/tex}So,\xa0{tex}Q \\rightarrow ( 0,5 ){/tex}and, R divides AB in the ratio 3 : 1{tex}\\therefore x _ { 2 } = \\frac { ( 3 ) ( 2 ) + ( 1 ) ( - 2 ) } { 3 + 1 }{/tex}{tex}= \\frac { 6 - 2 } { 4 } = \\frac { 4 } { 4 } = 1{/tex}{tex}y _ { 3 } = \\frac { ( 3 ) ( 8 ) + ( 1 ) ( 2 ) } { 3 + 1 }{/tex}{tex}= \\frac { 24 + 2 } { 4 } = \\frac { 26 } { 4 } = \\frac { 13 } { 2 }{/tex}So,\xa0{tex}\\mathrm { R } \\rightarrow \\left( 1 , \\frac { 13 } { 2 } \\right){/tex}

Find the coordinate of point which divide linesegment to (-2., 2)(2, 8)