The point which divides the line segment joining the points A(3, -2) and B(6, 7) internally in the ratio 3 : 2 lies in which of the following quadrants?

1. I

2. II

3. III

4. IV

Let point P be (x, y).

Given,

m₁ : m₂ = 3 : 2

By section-formula,

(x, y) = $\Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Substituting values we get :

$\Rightarrow (x, y) = \Big(\dfrac{3 \times 6 + 2 \times 3}{3 + 2}, \dfrac{3 \times 7 + 2 \times (-2)}{3 + 2}\Big) \\[1em] = \Big(\dfrac{18 + 6}{5}, \dfrac{21 - 4}{5}\Big) \\[1em] = \Big(\dfrac{24}{5}, \dfrac{17}{5}\Big).$

Here, both x and y are positive.

Therefore, the point lies in the 1st Quadrant.

Hence, Option 1 is the correct option.