1. ## coordinate geomatry

Determine the ratio in which the line 3X+Y-9=0 divides the line segment joining the points(1,3) and (2,7)

2. Hello, nitinnitin2!

Determine the ratio in which the line $3x+y-9\:=\:0$ divides
the line segment joining the points: $A(1,3)\text{ and }B(2,7).$

The line through $A$ and $B$ has slope: . $m \:=\:\frac{7-3}{2-1} \:=\:4$

. . and has the equation: . $y - 3 \:=\:4(x-1) \quad\Rightarrow\quad y \:=\:4x-1$

The other line is: . $y \:=\:-3x + 9$

They intersect when: . $4x - 1 \:=\:-3x+9 \quad\Rightarrow\quad x \:=\:\frac{10}{7} \quad\Rightarrow\quad y \:=\:\frac{33}{7}$

We have this diagram: . . $\begin{array}{ccccc}A & & P & & B \\
* & --- & * & --- & * \\
(1,3) & & \left(\tfrac{10}{7},\tfrac{33}{7}\right) & & (2,7) \end{array}$

. . $\overline{AP} \;=\;\sqrt{\left(\tfrac{10}{7}-1\right)^2 + \left(\tfrac{33}{7}-3\right)^2} \;=\;\sqrt{\left(\tfrac{3}{7}\right)^2 + \left(\tfrac{12}{7}\right)^2} \;=\; \sqrt{\tfrac{153}{49}} \:=\:\tfrac{3}{7}\sqrt{17}$

. . $\overline{PB} \;=\;\sqrt{\left(2-\tfrac{10}{7}\right)^2 + \left(7-\tfrac{33}{7}\right)^2} \;=\;\sqrt{\left(\tfrac{4}{7}\right)^2 + \left(\tfrac{16}{7}\right)^2} \;=\; \sqrt{\tfrac{272}{49}} \;=\;\tfrac{4}{7}\sqrt{17}$

Therefore: . $\overline{AP} : \overline{PB} \;\;=\;\;\tfrac{3}{7}\sqrt{17} : \tfrac{4}{7}\sqrt{17} \;\;=\;\;\boxed{ 3 : 4}$