Question

By using graph paper find the coordinates of a point which divides the line segment internally.

Answer 1

REF.Image

Let $(x_1,y_2)$ and $(x_2,y_2)$ be the Cartesian co-ordinates of the points

P and Q resp preferred to rectangle co-ordinate axis $\overline{OX}$ and

$\overline{OY}$ and the point R divides the line segment $\overline{PQ}$ interacually in a

given ratio $m:n$ i.e., $\overline{PR}:\overline{RQ}=m:n$

Let $(x,y)$ be the required co-ordinate of R. From P,Q and R draw

$\overline{PL},\overline{QN},\overline{RN}$ perpendicular or $\overline{OX}$ to cut $\overline{RN}$ at S and $\overline{QM}$ at T.

Thus,

$\overline{PS}=\overline{LN}=\overline{ON}-\overline{OL}=x-\overline{x_1}$

$\overline{PT}=LN=\overline{OM}-\overline{OL}=x_2-\overline{x_1}$

$\overline{RS}=\overline{RN}-\overline{SN}=\overline{RN}-\overline{PL}=y=\overline{y_1}$

and $\overline{QT}=\overline{QN}-\overline{TN}=\overline{QN}-\overline{PL}=y_2-y_1$

Again, $\overline{PR}/\overline{RQ}=m/n$

$\overline{RQ}/\overline{PR}+1=n/m+1$

$(\overline{RQ}+\overline{PR}/\overline{PR})=(m+n)/m$

$\overline{PQ}/\overline{PR}=(m+n)/m$

$\overline{PS}/\overline{PT}=\overline{RS}/\overline{QT}=\overline{PR}/\overline{PQ}$

Take $\overline{PS}/\overline{PT}=\overline{PR}/\overline{PQ}$

$(x-x_1)/(x_2-x_1)=m(m+n)$

$x=(m{x}_2+n{x}_1)/(m+n)$

Again Take, $\overline{RS}/\overline{QT}=\overline{PR}/\overline{RQ}$

$\therefore \left[\right(m{x}_2+n{x}_1\left)\right(m+n),(m{y}_2+n{y}_1\left)\right(m+n\left)\right]$