The ratio in which the x-axis divides the segment joining (3, 6) and (12, −3) is
Question:

The ratio in which the x-axis divides the segment joining (3, 6) and (12, −3) is

(a) 2: 1

(b) 1 : 2

(c) −2 : 1

(d) 1 : −2

Solution:

Let $P(x, 0)$ be the point of intersection of $x$-axis with the line segment joining $A(3,6)$ and $B(12,-3)$ which divides the line segment $A B$ in the ratio $\lambda: 1$.

Now according to the section formula if point a point $P$ divides a line segment joining $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ in the ratio $m$ : $n$ internally than,

$\mathrm{P}(x, y)=\left(\frac{m x_{1}+m x_{2}}{m+n}, \frac{n y_{1}+m y_{2}}{m+n}\right)$

Now we will use section formula as,

$(x, 0)=\left(\frac{12 \lambda+3}{\lambda+1}, \frac{-3 \lambda+6}{\lambda+1}\right)$

Now equate the y component on both the sides,

$\frac{-3 \lambda+6}{\lambda+1}=0$

On further simplification,

$\lambda=\frac{2}{1}$

So $x$-axis divides $A B$ in the ratio $\frac{2}{1}$