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

The correct option is A: $\frac{2}{7}$

Let us consider a point P$(-1,6)$ (refer figure given below) that divides A$(-3,10)$ and B$(6,-8)$ in that ratio $k:1$, then the coordinates of P are
$P(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$
Where $(x,y)=(-1,6),m:n=k:1,(x_1,y_1)=(-3,10)$ and $(x_2,y_2)=(6,-8)$
$\Rightarrow (-1,6)=(\frac{6\times k+(-3\times 1)}{k+1},\frac{(-8\times k)+(10\times 1)}{k+1})$

$\Rightarrow (-1,6)=(\frac{6k-3}{k+1},\frac{-8k+10}{k+1})\text{( Using section formula)}$

$\therefore \left(\frac{6k-3}{k+1}\right)=-1$

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

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

$\Rightarrow 7k=2$

$\therefore k=\frac{2}{7}$
Hence, the ratio in which line segment joining the points $(-3,10)$ and $(6,-8)$ is divided by $(-1,6)$ is $\frac{2}{7}$.