In what ratio is the segment joining the points (2,5) and (­6,­10) divided by the y­axis?

Using section formula, the coordinates of point that divides line joining A = $$(x_1 , y_1)$$ and B = $$(x_2 , y_2)$$ in the ratio a : b

= $$(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$$

Let the ratio in which the segment joining (2,5) and (6,10) divided by the y-axis = $$k$$ : $$1$$

Since, the line segment is divided by y-axis, thus x coordinate of the point will be zero, let the point of intersection = $$(0,y)$$

Now, point P (0,y) divides (2,5) and (6,10) in ratio = k : 1

=> $$0 = \frac{(6 \times k) + (2 \times 1)}{k + 1}$$

=> $$6k + 2 = 0$$

=> $$k = \frac{-2}{6} = \frac{-1}{3}$$

$$\therefore$$ Line segment joining (2,­5) and (­6,10) is divided by the Y ­axis in the ratio = 1 : 3 externally

=> Ans - (B)