Question

Find the ratio in which the point P$\left(\frac{3}{4},\frac{5}{12}\right)$ divides the line segment joining the points A$\left(\frac{1}{2},\frac{3}{2}\right)$ and B$\left(2,-5\right)$. [CBSE 2015]

Answer 1

Suppose P$\left(\frac{3}{4},\frac{5}{12}\right)$ divides the line segment joining the points A$\left(\frac{1}{2},\frac{3}{2}\right)$ and B$\left(2,-5\right)$ in the ratio k : 1.

Using section formula, we get

Coordinates of P = $\left(\frac{2k+{\displaystyle \frac{1}{2}}}{k+1},\frac{-5k+{\displaystyle \frac{3}{2}}}{k+1}\right)$

$\therefore \left(\frac{2k+{\displaystyle \frac{1}{2}}}{k+1},\frac{-5k+{\displaystyle \frac{3}{2}}}{k+1}\right)=\left(\frac{3}{4},\frac{5}{12}\right)$

$\Rightarrow \frac{2k+{\displaystyle \frac{1}{2}}}{k+1}=\frac{3}{4}$ and $\frac{-5k+{\displaystyle \frac{3}{2}}}{k+1}=\frac{5}{12}$

Now,

$$\begin{gathered} \frac{2k+{\displaystyle \frac{1}{2}}}{k+1}=\frac{3}{4} \\ \Rightarrow 8k+2=3k+3 \\ \Rightarrow 5k=1 \\ \Rightarrow k=\frac{1}{5} \end{gathered}$$

Putting k = $\frac{1}{5}$ in $\frac{-5k+{\displaystyle \frac{3}{2}}}{k+1}=\frac{5}{12}$, we get

LHS = $\frac{-5\times {\displaystyle \frac{1}{5}}+{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{1}{5}}+1}=\frac{-1+{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{1}{5}}+1}=\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{6}{5}}}=\frac{5}{12}$ = RHS

Thus, the required ratio is $\frac{1}{5}$ : 1 or 1 : 5.