Question

In what ratio does the line $$x - 5y + 15 = 0$$ divide the join of $$\mathrm{A}(2,1)$$ and $$\mathrm{B}(-3, 6)$$? Also, find the co-ordinates of their point of intersection. 

[Hint. Let the given line divide AB in the ratio k : 1 at point P. Then, the co-ordinates of P would satisfy x – 5y + 15 = 0. Find k.]

Answer 1

Let the line divide the segment joining $$A(2, 1)$$ and $$B(-3, 6)$$ in the ratio $$k : 1$$ at point $$P$$.

Using the section formula, the coordinates of $$P$$ are:

$$P = \left(\frac{k(-3) + 1(2)}{k + 1}, \frac{k(6) + 1(1)}{k + 1}\right)$$

$$= \left(\frac{-3k + 2}{k + 1}, \frac{6k + 1}{k + 1}\right)$$

Since $$P$$ lies on the line $$x - 5y + 15 = 0$$: 

$$\left(\frac{-3k + 2}{k + 1}\right) - 5\left(\frac{6k + 1}{k + 1}\right) + 15 = 0$$

Multiplying throughout by $$(k + 1)$$:

$$(-3k + 2) - 5(6k + 1) + 15(k + 1) = 0$$

$$-3k + 2 - 30k - 5 + 15k + 15 = 0$$

$$-18k + 12 = 0$$

$$18k = 12$$

$$\implies k = \frac{12}{18} = \frac{2}{3}$$

Thus, the line divides the join of $$AB$$ in the ratio $$2 : 3$$.

Now, substitute $$k = \frac{2}{3}$$ into the coordinates of $$P$$:

$$x = \frac{-3\left(\frac{2}{3}\right) + 2}{\frac{2}{3} + 1}$$

$$= \frac{-2 + 2}{\frac{5}{3}}$$

$$= 0$$

$$y = \frac{6\left(\frac{2}{3}\right) + 1}{\frac{2}{3} + 1}$$

$$= \frac{4 + 1}{\frac{5}{3}}$$

$$= \frac{5}{\frac{5}{3}}$$

$$= 3$$

Therefore, the coordinates of the point of intersection are $$(0, 3)$$.