(i) In what ratio is the line segment joining the points (−2,−3) and (3, 7)

Solution:

(i) The ratio in which the $y$-axis divides two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is $\lambda: 1$

The co-ordinates of the point dividing two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ in the ratio $m: n$ is given as,

$(x, y)=\left(\left(\frac{\lambda x_{2}+x_{1}}{\lambda+1}\right),\left(\frac{\lambda y_{2}+y_{1}}{\lambda+1}\right)\right) ;$ where $\lambda=\frac{m}{n}$

Here the two given points are A(−2,−3) and B(3,7).

Since, the point is on the y-axis so, x coordinate is 0. 

$\frac{3 \lambda-2}{1}=0$

$\Rightarrow \lambda=\frac{2}{3}$

Thus the given points are divided by the $y$-axis in the ratio $2: 3$.

 

The co-ordinates of this point $(x, y)$ can be found by using the earlier mentioned formula.

$(x, y)=\left(\left(\frac{\frac{2}{3}(3)+(-2)}{\frac{2}{3}+1}\right),\left(\frac{\frac{2}{3}(7)+(-3)}{\frac{2}{3}+1}\right)\right)$

$(x, y)=\left(\left(\frac{\frac{6-2(3)}{3}}{\frac{2+3}{3}}\right),\left(\frac{\frac{14-3(3)}{3}}{\frac{2+3}{3}}\right)\right)$

$(x, y)=\left(\left(\frac{0}{5}\right),\left(\frac{5}{5}\right)\right)$

$(x, y)=(0,1)$

Thus the co-ordinates of the point which divides the given points in the required ratio are $(0,1)$.

(ii) The co-ordinates of a point which divided two points  and  internally in the ratio  is given by the formula,

$(x, y)=\left(\left(\frac{m x_{2}+n x_{1}}{m+n}\right),\left(\frac{m y_{2}+n y_{1}}{m+n}\right)\right)$

Here it is said that the point $\left(-5,-\frac{21}{5}\right)$ divides the points $(-3,-1)$ and $(-8,-9)$. Substituting these values in the above formula we have,

$\left(-5,-\frac{21}{5}\right)=\left(\left(\frac{m(-8)+n(-3)}{m+n}\right),\left(\frac{m(-9)+n(-1)}{m+n}\right)\right)$

Equating the individual components we have,

$-5=\frac{m(-8)+n(-3)}{m+n}$

$-5 m-5 n=-8 m-3 n$

$3 m=2 n$

$\frac{m}{n}=\frac{2}{3}$

Therefore the ratio in which the line is divided is $2: 3$.