Find the coordinates of the point $ R $ on the line segment joining the points $ \mathrm{P}(-1,3) $ and $ \mathrm{Q}(2,5) $ such that $ \mathrm{PR}=\frac{3}{5} \mathrm{PQ} $.

Given:

The line segment joining the points \( \mathrm{P}(-1,3) \) and \( \mathrm{Q}(2,5) \) such that \( \mathrm{PR}=\frac{3}{5} \mathrm{PQ} \).

To do:

We have to find the coordinates of the point \( R \).

Solution:

Let $R(x, y)$ be the point which divides the line segment joining the points $P(-1,3)$ and $Q(2,5)$

Given,

$P R=\frac{3}{5} P Q$

$\frac{P Q}{P R}=\frac{5}{3}$

$\frac{P R+R Q}{P R}=\frac{5}{3}$           (Since $PQ=PR+RQ$)

$1+\frac{R Q}{P R} =\frac{5}{3}$

$\frac{RQ}{P R}=\frac{5}{3}-1$

$=\frac{5-3}{3}$

$=\frac{2}{3}$

This implies,

$R Q: P R =2: 3$

$P R: R Q=3: 2$

Therefore,

$R(x, y)$ divides the line segment joining the points $P(-1,3)$ and $Q(2,5)$ in the ratio $3: 2$.

Using section formula, we get,

$(x,y)=(\frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}ny_{1}}{m+n})$

$(x, y)=(\frac{3(2)+2(-1)}{3+2}, \frac{3(5)+2(3)}{3+2})$

$=(\frac{6-2}{5}, \frac{15+6}{5})$

$=(\frac{4}{5}, \frac{21}{5})$

Therefore, the coordinates of the point \( R \) are $(\frac{4}{5}, \frac{21}{5})$.

Tutorialspoint

Updated on: 10-Oct-2022

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