Question

In what ratio is the line segment joining the points A(−2, −3) and B(3, 7) divided by the y-axis? Also, find the coordinates of the points of division.

Answer 1

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are
$P\left(\frac{3k-2}{k+1},\frac{7k-3}{k+1}\right)$
But P lies on the y-axis; so, its abscissa is 0.
$$\begin{gathered} \mathrm{Therefore},\frac{3k-2}{k+1}=0 \\ \Rightarrow 3k-2=0\Rightarrow 3k=2\Rightarrow k=\frac{2}{3}\Rightarrow k=\frac{2}{3} \end{gathered}$$
Therefore, the required ratio is $\frac{2}{3}$ : 1, which is same as 2 : 3.
Thus, the x-axis divides the line AB in the ratio 2:3 at the point P.
Applying k=$\frac{2}{3}$, we get the coordinates of point P:
$$\begin{gathered} P\left(0,\frac{7k-3}{k+1}\right) \\ =P\left(0,\frac{7\times {\displaystyle \frac{2}{3}}-3}{{\displaystyle \frac{2}{3}}+1}\right) \\ =P\left(0,\frac{{\displaystyle \frac{14-9}{3}}}{{\displaystyle \frac{2+3}{3}}}\right) \\ =P\left(0,\frac{5}{5}\right) \\ =P\left(0,1\right) \end{gathered}$$
Hence, the point of intersection of AB and the x-axis is P(0, 1).