Question

In what ratio is the line segment joining A(2, −3) and B(5, 6) divided by the x-axis? Also, find the coordinates of the point 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
$P=\left(\frac{5k+2}{k+1},\frac{6k-3}{k+1}\right)$
But P lies on the x-axis; so, its ordinate is 0.
$$\begin{gathered} \mathrm{Therefore},\frac{6k-3}{k+1}=0 \\ \Rightarrow 6k-3=0\Rightarrow 6k=3\Rightarrow k=\frac{3}{6}\Rightarrow k=\frac{1}{2} \end{gathered}$$
Therefore, the required ratio is $\frac{1}{2}$ : 1, which is same as 1 : 2.
Thus, the x-axis divides the line AB in the ratio 1 : 2 at the point P.
Applying k = $\frac{1}{2}$, we get the coordinates of point :
$$\begin{gathered} P\left(\frac{5k+2}{k+1},0\right) \\ =P\left(\frac{5\times \frac{1}{2}+2}{\frac{1}{2}+1},0\right) \\ =P\left(\frac{{\displaystyle \frac{5+4}{2}}}{{\displaystyle \frac{1+2}{2}}},0\right) \\ =P\left(\frac{9}{3},0\right) \\ =P\left(3,0\right) \end{gathered}$$
Hence, the point of intersection of AB and the x-axis is P(3, 0).