Question

If the line 3x + y - 9 = 0 divides the segment joining the points (1, 3) and (2, 7) in the ratio a : b , Find a + b

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 thesection 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 $a:b=3:4$ internally.

So $a+b=7$