Points P, Q and Richard in …

Points P, Q and R in order divide a line segment joining the points A(1, 6) and B(5, -2) in 4 equal parts.

P divides AB in the ratio of 1:3 Let coordinates of P be (x, y), then
{tex}x = \frac { m x _ { 2 } + n x _ { 1 } } { m + n } = \frac { 1 \times 5 + 3 \times 1 } { 1 + 3 }{/tex}
{tex}= \frac { 5 + 3 } { 4 } = \frac { 8 } { 4 } = 2{/tex}
{tex}y = \frac { m y _ { 2 } + n y _ { 1 } } { m + n } = \frac { 1 \times ( - 2 ) + 3 \times 6 } { 1 + 3 }{/tex}
{tex}= \frac { - 2 + 18 } { 4 } = \frac { 16 } { 4 } = 4{/tex}
{tex}\therefore{/tex} Coordinates of P are (2, 4)
Similarly,
Q divides AB in 2:2 or 1:1 and Q is midpoint of AB.
{tex}\therefore{/tex} Coordinates of Q will be {tex}\left( \frac { 1 + 5 } { 2 } , \frac { 6 - 2 } { 2 } \right){/tex}
or {tex}\left( \frac { 6 } { 2 } , \frac { 4 } { 2 } \right){/tex} or (3, 2)
and R divides AB in the ratio of 3:1
Coordinates of R will be 
{tex}\left( \frac { 3 \times 5 + 1 \times 1 } { 3 + 1 } , \frac { 3 \times ( - 2 ) + 1 \times 6 } { 3 + 1 } \right){/tex}
or {tex}\left( \frac { 15 + 1 } { 4 } , \frac { - 6 + 6 } { 4 } \right) \text { or } \left( \frac { 16 } { 4 } , \frac { 0 } { 4 } \right) \text { or } ( 4,0 ){/tex}