Find the ratio in which the point P(−1, y) lying on the line segment

Suppose P(−1, y) divides the line segment joining A(−3, 10) and B(6 −8) in the ratio k : 1.

Using section formula, we get

Coordinates of $\mathrm{P}=\left(\frac{6 k-3}{k+1}, \frac{-8 k+10}{k+1}\right)$

$\therefore\left(\frac{6 k-3}{k+1}, \frac{-8 k+10}{k+1}\right)=(-1, y)$

$\Rightarrow \frac{6 k-3}{k+1}=-1$ and $y=\frac{-8 k+10}{k+1}$

Now,

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

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

$\Rightarrow 7 k=2$

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

So, P divides the line segment AB in the ratio 2 : 7.

Putting $k=\frac{2}{7}$ in $y=\frac{-8 k+10}{k+1}$, we get

$y=\frac{-8 \times \frac{2}{7}+10}{\frac{2}{7}+1}=\frac{-16+70}{2+7}=\frac{54}{9}=6$

Hence, the value of y is 6.