Question

Find the coordinates of the points which divide the line segment joining $A(-2,2)$ and $B(2,8)$ into four equals parts.

Answer 1

We know that by section formula, the co-ordinates of the points which divide internally the line segment joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$ is

$(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Let $P$, $Q$ and $R$ be the points that divide $AB$ in four equal parts.

Let $AP,PQ,QR,RB,k$ all be equal to $k.$

So,

$\frac{AQ}{QB}=\frac{AP+PQ}{QR+RB}$

$=\frac{k+k}{k+k}$

$=\frac{2k}{2k}$

$=1:1$

Therefore, $Q$ divides $AB$ in two equal parts, $AQ$ and $QB$.

So, the coordinates of $Q$ is,

$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{-2+2}{2},\frac{2+8}{2}\right)$

$=\left(0,5\right)$

Thus, $C_2=\left(0,5\right)$.

Similarly,

$\frac{AP}{PQ}=\frac{k}{k}$

$=1:1$

So, point $P$ divides $AQ$ in two equal parts.

The coordinates of $P$ are,

$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{-2+0}{2},\frac{2+5}{2}\right)$

$=\left(-1,\frac{7}{2}\right)$

Thus, $C_1=\left(-1,\frac{7}{2}\right)$.

Also, in the same way,

$\frac{QR}{RB}=\frac{k}{k}$

$=1:1$

So, point $R$ divides $QB$ in two equal parts.

The coordinates of $R$ are,

$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{0+2}{2},\frac{5+8}{2}\right)$

$=\left(1,\frac{13}{2}\right)$

Thus, $C_3=\left(1,\frac{13}{2}\right)$.

Therefore, $C_1=\left(-1,\frac{7}{2}\right)$, $C_2=\left(0,5\right)$ and $C_3=\left(1,\frac{13}{2}\right)$.