Question

The mid-point P of the line segment joining the points A(−10, 4) and B(−2, 0) lies on the line segment joining the points C(−9, −4) and D(−4, y). Find the ratio in which P divides CD. Also, find the value of y. [CBSE 2014]

Answer 1

It is given that P is the mid-point of the line segment joining the points A(−10, 4) and B(−2, 0).

∴ Coordinates of P = $\left(\frac{-10+\left(-2\right)}{2},\frac{4+0}{2}\right)=\left(\frac{-12}{2},\frac{4}{2}\right)=\left(-6,2\right)$

Suppose P divides the line segment joining the points C(−9, −4) and D(−4, y) in the ratio k : 1.

Using section formula, we get

Coordinates of P = $\left(\frac{-4k-9}{k+1},\frac{ky-4}{k+1}\right)$

$$\begin{gathered} \therefore \left(\frac{-4k-9}{k+1},\frac{ky-4}{k+1}\right)=\left(-6,2\right) \\ \Rightarrow \frac{-4k-9}{k+1}=-6\mathrm{and}\frac{ky-4}{k+1}=2 \end{gathered}$$

Now,

$$\begin{gathered} \frac{-4k-9}{k+1}=-6 \\ \Rightarrow -4k-9=-6k-6 \\ \Rightarrow 2k=3 \\ \Rightarrow k=\frac{3}{2} \end{gathered}$$

So, P divides the line segment CD in the ratio 3 : 2.

Putting k = $\frac{3}{2}$ in $\frac{ky-4}{k+1}=2$, we get
$$\begin{gathered} \frac{{\displaystyle \frac{3y}{2}}-4}{{\displaystyle \frac{3}{2}}+1}=2 \\ \Rightarrow \frac{3y-8}{5}=2 \\ \Rightarrow 3y-8=10 \\ \Rightarrow 3y=18 \\ \Rightarrow y=6 \end{gathered}$$

Hence, the value of y is 6.