Find the coordinate of the point …

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} 
{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, {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, {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, {tex}\mathrm { R } \rightarrow \left( 1 , \frac { 13 } { 2 } \right){/tex}