Suppose M = (1, 1), N = (-1,3), S = (2, 7), T = (0,-4). If A and B, respectively, divide MN and ST in the ratio 2 : 3, what is the equation of line AB?
Solution
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 $$y=\frac{4}{5}\cdot x+\frac{41}{25}=>20x-25y+41=0$$
