Question

Find the ratio in which the line segment joining the points $(-3,10)$ and $(6-8)$ is divided by $(-1,6)$

Answer 1

Consider the points be $A\left(-3,10\right),B\left(6,-8\right),C\left(-1,6\right)$

And we need to find the ratio between $AC$ and $CB$

So, let the ratio be $k:1$

Therefore,

$m_1=k,m_2=1$

$x_1=-3,y_1=10$

$x_2=6,y_2=-8$

And $x=-1,y=6$

Now, using section formula

$x=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}$

$\Rightarrow -1=\frac{k\times 6+1\times -3}{k+1}$

$\Rightarrow -1=\frac{6k-1}{k+1}$

$\Rightarrow -1\left(k+1\right)=6k-3$

$\Rightarrow -k-1=6k-3$

$\Rightarrow -k-6k=-3+1$

$\Rightarrow -7k=-2$

$\Rightarrow k=\frac{-2}{7}$

$\Rightarrow k=\frac{2}{7}$

Therefore, the ratio is $k:1$

$=\frac{2}{7}:1$

Now multiply $7$ both sides

$=7\times \frac{2}{7}:7\times 1$

$=2:7$

Hence, The Ratio is $2:7$