Question
Find the coordinates of the points which divide the line segment joining $A(-2, 2)$ and $B(2, 8)$ into four equal parts.
✓
Answer
Let $P\left(x_1, y_1\right) Q\left(x_2, y_2\right)$ and $R\left(x_3, y_3\right)$ be the points which divide the line segment $AB$ into four equal parts.
Then, $P$ divides $AB$ in the ratio $1 : 3$ internally.
$x=\frac{mx_2+nx_1}{m+n}$
$\therefore x _ { 1 } = \frac { ( 1 ) ( 2 ) + ( 3 ) ( - 2 ) } { 1 + 3 }$
$= \frac { 2 - 6 } { 4 } = - \frac { 4 } { 4 } = - 1$
$y=\frac{my_2+ny_1}{m+n}$
$y _ { 1 } = \frac { ( 1 ) ( 8 ) + ( 3 ) ( 2 ) } { 1 + 3 }$
$= \frac { 8 + 6 } { 4 } = \frac { 14 } { 4 } = \frac { 7 } { 2 }$
So, $\mathrm { P } \rightarrow \left( - 1 , \frac { 7 } { 2 } \right)$
Also, $Q$ divides $AB$ in the ratio $1 : 1$ i.e.
$Q$ is the mid point of $AB$
$x _ { 2 } = \frac { - 2 + 2 } { 2 } = 0$
$y _ { 2 } = \frac { 2 + 8 } { 2 } = \frac { 10 } { 2 } = 5$
So, $Q \rightarrow ( 0,5 )$
and, $R$ divides $AB$ in the ratio $3 : 1$
$\therefore x _ { 2 } = \frac { ( 3 ) ( 2 ) + ( 1 ) ( - 2 ) } { 3 + 1 }$
$= \frac { 6 - 2 } { 4 } = \frac { 4 } { 4 } = 1$
$y _ { 3 } = \frac { ( 3 ) ( 8 ) + ( 1 ) ( 2 ) } { 3 + 1 }$
$= \frac { 24 + 2 } { 4 } = \frac { 26 } { 4 } = \frac { 13 } { 2 }$
So, $\mathrm { R } \rightarrow \left( 1 , \frac { 13 } { 2 } \right)$
Then, $P$ divides $AB$ in the ratio $1 : 3$ internally.
$x=\frac{mx_2+nx_1}{m+n}$
$\therefore x _ { 1 } = \frac { ( 1 ) ( 2 ) + ( 3 ) ( - 2 ) } { 1 + 3 }$
$= \frac { 2 - 6 } { 4 } = - \frac { 4 } { 4 } = - 1$
$y=\frac{my_2+ny_1}{m+n}$
$y _ { 1 } = \frac { ( 1 ) ( 8 ) + ( 3 ) ( 2 ) } { 1 + 3 }$
$= \frac { 8 + 6 } { 4 } = \frac { 14 } { 4 } = \frac { 7 } { 2 }$
So, $\mathrm { P } \rightarrow \left( - 1 , \frac { 7 } { 2 } \right)$
Also, $Q$ divides $AB$ in the ratio $1 : 1$ i.e.
$Q$ is the mid point of $AB$
$x _ { 2 } = \frac { - 2 + 2 } { 2 } = 0$
$y _ { 2 } = \frac { 2 + 8 } { 2 } = \frac { 10 } { 2 } = 5$
So, $Q \rightarrow ( 0,5 )$
and, $R$ divides $AB$ in the ratio $3 : 1$
$\therefore x _ { 2 } = \frac { ( 3 ) ( 2 ) + ( 1 ) ( - 2 ) } { 3 + 1 }$
$= \frac { 6 - 2 } { 4 } = \frac { 4 } { 4 } = 1$
$y _ { 3 } = \frac { ( 3 ) ( 8 ) + ( 1 ) ( 2 ) } { 3 + 1 }$
$= \frac { 24 + 2 } { 4 } = \frac { 26 } { 4 } = \frac { 13 } { 2 }$
So, $\mathrm { R } \rightarrow \left( 1 , \frac { 13 } { 2 } \right)$
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