Question

Find the ratio in which the point P(−1, y) lying on the line segment joining A(−3, 10) and B(6 −8) divides it. Also find the value of y. [CBSE 2013]

Answer 1

Suppose P(−1, y) divides the line segment joining A(−3, 10) and B(6 −8) in the ratio k : 1.

Using section formula, we get

Coordinates of P = $\left(\frac{6k-3}{k+1},\frac{-8k+10}{k+1}\right)$

$\therefore \left(\frac{6k-3}{k+1},\frac{-8k+10}{k+1}\right)=\left(-1,y\right)$

$\Rightarrow \frac{6k-3}{k+1}=-1$ and $y=\frac{-8k+10}{k+1}$

Now,

$$\begin{gathered} \frac{6k-3}{k+1}=-1 \\ \Rightarrow 6k-3=-k-1 \\ \Rightarrow 7k=2 \\ \Rightarrow k=\frac{2}{7} \end{gathered}$$

So, P divides the line segment AB in the ratio 2 : 7.

Putting k = $\frac{2}{7}$ in $y=\frac{-8k+10}{k+1}$, we get

$y=\frac{-8\times {\displaystyle \frac{2}{7}}+10}{{\displaystyle \frac{2}{7}}+1}=\frac{-16+70}{2+7}=\frac{54}{9}=6$

Hence, the value of y is 6.