If the points $P, Q (x, 7), R, S (6, y)$ in this order divide the line segment joining $A (2, p)$ and $B (7,10)$ in 5 equal parts, find $x, y$ and $p$.

Given:

Points $P, Q (x, 7), R, S (6, y)$ in this order divide the line segment joining $A (2, p)$ and $B (7,10)$ in 5 equal parts.

To do:

We have to find $x, y$ and $p$.

Solution:

Points $P, Q (x, 7), R, S (6, y)$ in order divide a line segment joining $A (2, p)$ and $B (7, 10)$ in 5 equal parts.

This implies,

$AP = PQ = QR = RS = SB$
$Q$ is the mid-point of $A$ and $S$

Using mid-point formula, we get,

\( x=\frac{2+6}{2} \)

\( =\frac{8}{2} \)

\( =4 \)
\( 7=\frac{y+p}{2} \)

\( \Rightarrow y+p=14 \)......(i)
\( \mathrm{S} \) divides \( \mathrm{QB} \) in the ratio $2: 1$.

Using the section formula, if a point $( x,\ y)$ divides the line joining the points $( x_1,\ y_1)$ and $( x_2,\ y_2)$ in the ratio $m:n$, then 

$(x,\ y)=( \frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n})$

This implies,

\( y=\frac{2 \times 10+1 \times 7}{2+1} \)

\( =\frac{20+7}{3} \)

\( =\frac{27}{3} \)

\( =9 \)
This implies,

\( 9+p=14 \)

\( p=14-9 \)

\( p=5 \)

The values of $x, y$ and $p$ are $4, 9$ and $5$ respectively.