Question

If the ratio in which the line $3x+y-9=0$ divides the segment joining the points $(1,3)$ and $(2,7)$ is $p:q$, where $p$ and $q$ are co-primes, then $p+q=?$

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 $(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$
Let the ratio be $k:1$

Substituting $(x_1,y_1)=(1,3)$ and $(x_2,y_2)=(2,7)$ in the
section formula, we get the point which divides as $(\frac{k\left(2\right)+1\left(1\right)}{k+1},\frac{k\left(7\right)+1\left(3\right)}{k+1})=(\frac{2k+1}{k+1},\frac{7k+3}{k+1})$ Since this point lies on the line $3x+y-9=0$, we have

$3\left(\frac{2k+1}{k+1}\right)+\frac{7k+3}{k+1}-9=0$
$=>6k+3+7k+3-9k-9=0$

$4k-3=0$

$k=\frac{3}{4}$

Hence, the ratio is $3:4$ internally.

Thus $p:q=3:4$

And therefore $p+q=3+4=7$