Question

The ratio in which the $x$ axis divides the line segment joining the points $(2,-3)$ and $(5,6)$ is $m:n$, where $m$ and $n$ are co-primes, find the value of $m+n$.

Answer 1

Using the section formula, if a point $(x,y)$ divides the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$, then $(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Substituting $(x_1,y_1)=(2,-3)$ and $(x_2,y_2)=(5,6)$ in the section formula, we get the point:

$(\frac{m\left(5\right)+n\left(2\right)}{m+n},\frac{m\left(6\right)+n(-3)}{m+n})=(\frac{5m+2n}{m+n},\frac{6m-3n}{m+n})$

As the point lies on the $x$ - axis, the $y$ co-ordinate of the point $=0$
$⟹\frac{6m-3n}{m+n}=0$

$6m=3n$

$m:n=3:6=1:2$