**Question:**

The point which divides the line segment joining the points (7, – 6) and (3, 4) in ratio 1: 2 internally lies in the

(a) I quadrant

(b) II quadrant

(c) III quadrant

(d) IV quadrant

**Solution:**

(d) If P(x, y) divides the line segment joining A(x1,y2) and B(x2, y2) internally in the ratio

$m: n$, then $x=\frac{m x_{2}+n x_{1}}{m+n}$ and $y=\frac{m y_{2}+n y_{1}}{m+n}$

Given that, $x_{1}=7, y_{1}=-6, x_{2}=3, y_{2}=4, m=1$ and $n=2$

$\therefore$ $x=\frac{1(3)+2(7)}{1+2}, y=\frac{1(4)+2(-6)}{1+2}$ [by section formula]

$\Rightarrow$ $x=\frac{3+14}{3}, y=\frac{4-12}{3}$

$\Rightarrow$ $x=\frac{17}{3}, y=-\frac{8}{3}$

So, $(x, y)=\left(\frac{17}{3},-\frac{8}{3}\right)$ lies in IV quadrant.

[since, in IV quadrant, $x$-coordinate is positive and $y$-coordinate is negative]