# 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.

Updated on: 10-Oct-2022

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