Given the ratio and two points on the line, we can find out the point that divides the line in that ratio using internal section formulae $$\left(\frac{\left(mx_1+nx_2\right)}{m+n},\frac{\left(my_1+ny_2\right)}{m+n}\right)$$.

The divider of line MN(point A) is as follows.

$$\left(\frac{\left(3\cdot1+2\cdot-1\right)}{3+2},\frac{\left(3\cdot1+2\cdot3\right)}{3+2}\right)$$

A$$\left(\frac{1}{5},\frac{9}{5}\right)$$

Similarly, for line ST (point B)

$$\left(\frac{\left(3\cdot2+2\cdot0\right)}{3+2},\frac{\left(3\cdot7+2\cdot-4\right)}{3+2}\right)$$

B$$\left(\frac{6}{5},\frac{13}{5}\right)$$

Also, do the same for the other line; we get points A and B. Now, find the line equation using the formula y=mx+c.

m=$$\frac{y_2-y_1}{x_2-x_1}\ =\ \frac{\left(\frac{13}{5}-\frac{9}{5}\right)}{\frac{6}{5}-\frac{1}{5}}\ =\ \frac{4}{5}$$

The euation becoms $$y=\frac{4}{5}\cdot x+c$$

Substotue one of those two points to get c value.$$\frac{13}{5}=\frac{4}{5}\cdot\frac{6}{5}+c=>c=\frac{41}{25}$$

The equation will be $$y=\frac{4}{5}\cdot x+\frac{41}{25}=>20x-25y+41=0$$