In what ratio is the line joining the points $(1,3)$ and $(2,7)$ divided by the line $3x+y=9$?

Question

In what ratio the line, joining the points $(1,3)$ and $(2,7)$, divided by the line $3x+y=9$?

My Attempt:

Let the ratio be $m:n$. Then $$(x,y)=\left(\frac {mx_2+nx_1}{m+n} , \frac {my_2+ny_1}{m+n}\right)$$ $$=\left(\frac {m.2+n.1}{m+n}, \frac {m.7+n.3}{m+n}\right)$$ $$=\left(\frac {2m+n}{m+n}, \frac {7m+3n}{m+n}\right).$$

Answer 1

Alternatively, solve the line directly:

$$ \begin{cases} y=4x -1 \\ y = 9 - 3x \end{cases}$$

Thus we sovle for $x=\frac{10}{7}$. (The point is $(\frac{10}{7},\frac{33}{7})$, but you'll only need solve $x$ or solve $y$, no need for both.)

Notice that ratio wise, we could just focus on one of $x$ or $y$ axis, since it is a line (linear), focus on $x$-axis below:

$$\frac{m}{n}=\frac{\frac{10}{7}-1}{2-\frac{10}{7}}=\frac{3}{4}$$

EDIT

Segments over a line always have the same ratio for its projection onto axis or another line. For the following picture, we have

$$\frac{x_1}{y_1}=\frac{x_2}{y_2}, \text{ and thus } \frac{x_1}{x_2} = \frac{y_1}{y_2}$$

Answer 2

HINT: your first equation has the form $$y=mx+n$$ with $$m=\frac{7-3}{2-1}=4$$ thus we have $$y=4x+n$$ plugging one Point in this eqaution we have $$y=4x-1$$ can you compute the intersection Point with $$y=-3x+9$$?

Answer 3

Let $A(1,3)$, $B(2,7)$ and $C$ be our point such that $AC:CB=m:n$. Thus, $$C\left(\frac{n+2m}{m+n},\frac{3n+7m}{m+n}\right)$$ and $$3\cdot\frac{n+2m}{m+n}+\frac{3n+7m}{m+n}=9$$ or $$4m=3n,$$ which gives $m:n=3:4$.

Answer 4

The point diving the line segment $AB$, where $A=(1,3), B=(2,7)$, in the ratio $t:(1-t)$ is $$\big(2t+(1-t), \;7t+3(1-t)\big)=\big(t+1,\; 4t+3\big)$$ As this point also lies on the line $3x+y=9$, it must satisfy $$3(t+1)+(4t+3)=9\\ \Longrightarrow t=\frac 37\quad \Longrightarrow t:(1-t)=\color{red}{3:4}$$.