Question

The ratio in which the line joining $A(-4,2)$ and $B(3,6)$ is divided by point $P(x,3)$ is $k$. Then $9k=?$

Answer 1

Let $P(x,3)$ divide the line segment joining the points $A(-4,2)$ and $B(3,6)$ in the ratio $k:1$
$\therefore$ Coordinates of $P$ is $(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})=(\frac{3k-4}{k+1},\frac{6k+2}{k+1})$
But coordinate of $P$ is $(x,3)$
$\Rightarrow \frac{6k+2}{k+1}=3$
$6k+2=3k+3$
$3k=1\Rightarrow k=\frac{1}{3}$
$\therefore$ The required ratio is $\frac{1}{3}:1$ i.e., $1:3$ (internally)
$\therefore x=\frac{3k-4}{k+1}$
Putting $k=\frac{1}{3}$, we get
$x=\frac{3\times \frac{1}{3}-4}{\frac{1}{3}+1}=\frac{1-4}{\frac{1+3}{3}}=\frac{-3}{4/3}=\frac{-9}{4}$

$\therefore$ $9k=3$