Find the ratio in which y-axis divides the line segment joining the points A(5, –6) and B(–1, 4) Also,

Solution:

Section formula: if the point $(x, y)$ divides the line segment joining the points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ internally in the ratio $k: 1$, then the coordinates $(x, y)=\left(\frac{k x_{2}+x_{1}}{k+1}, \frac{k y_{2}+y_{1}}{k+1}\right)$

Let the point P(0, y) divides the line segment joining the points A(5, –6) and B(–1, 4) in the ratio k : 1.

Therefore, using section formula, the coordinates of P are:

$(0, y)=\left(\frac{k(-1)+1(5)}{k+1}, \frac{k(4)+1(-6)}{k+1}\right)$

$\Rightarrow(0, y)=\left(\frac{-k+5}{k+1}, \frac{4 k-6}{k+1}\right)$

$\Rightarrow 0=\frac{-k+5}{k+1}$ and $y=\frac{4 k-6}{k+1}$

$\Rightarrow 0=\frac{-k+5}{k+1}$

$\Rightarrow-k+5=0$

$\Rightarrow k=5$

Now, $y=\frac{4 k-6}{k+1}$

$\Rightarrow y=\frac{4(5)-6}{5+1} \quad(\because k=5)$

$\Rightarrow y=\frac{20-6}{6}$

$\Rightarrow y=\frac{14}{6}$

$\Rightarrow y=\frac{7}{3}$

Hence, the $y$-axis divides the line segment joining the points $A(5,-6)$ and $B(-1,4)$ in the ratio $5: 1$.

and the coordinates of the point of division are $\left(0, \frac{7}{3}\right)$.