$\mathrm{P}(-1, 4), \mathrm{Q}(11, -8)$ divide $\overline{\mathrm{AB}}$ hormonically in the ratio $3 : 2$ then $\mathrm{A}, \mathrm{B}$ are 

Let A=(a, b), B=(x, y)

Given points P=(-1, 4), Q=(11, -8)

Given that P and Q divides $\overline{\mathrm{AB}}$ hormonically in the ratio 3:2 then

$$\begin{gathered} \mathrm{P}=\left[\frac{3\left(\mathrm{x}\right)+2\left(\mathrm{a}\right)}{3+2}, \frac{3\left(\mathrm{y}\right)+2\left(\mathrm{b}\right)}{3+2}\right] \\ \left[-1, 4\right]=\left[\frac{3\mathrm{x}+2\mathrm{a}}{5}, \frac{3\mathrm{y}+2\mathrm{b}}{5}\right] \\ -1=\frac{3\mathrm{x}+2\mathrm{a}}{5}, 4=\frac{3\mathrm{y}+2\mathrm{b}}{5} \\ \Rightarrow 3\mathrm{x}+2\mathrm{a}=-5 \to \left(1\right), 3\mathrm{y}+2\mathrm{b}=20 \to \left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Now} \mathrm{Q}=\left[\frac{3\left(\mathrm{x}\right)-2\left(\mathrm{a}\right)}{3-2}, \frac{3\left(\mathrm{y}\right)-2\left(\mathrm{b}\right)}{3-2}\right] \\ \Rightarrow \left[11, 8\right]=\left[\frac{3\mathrm{x}-2\mathrm{a}}{1}, \frac{3\mathrm{y}-2\mathrm{b}}{1}\right] \\ \Rightarrow 3\mathrm{x}-2\mathrm{a}=11 \to \left(3\right) \\ \mathrm{and} 3\mathrm{y}-2\mathrm{b}=-8 \to \left(4\right) \\ \left(1\right)+\left(3\right)\Rightarrow 6\mathrm{x}=6\Rightarrow \mathrm{x}=1 \\ \left(2\right)+\left(4\right)\Rightarrow 6\mathrm{y}=12\Rightarrow \mathrm{y}=2 \\ \mathrm{from} \left(1\right) 3\left(1\right)+2\mathrm{a}=-5 \\ \Rightarrow 3+2\mathrm{a}=-5 \\ \Rightarrow 2\mathrm{a}=-8 \\ \Rightarrow \mathrm{a}=-4 \\ \mathrm{from} \left(2\right) \Rightarrow 3\left(2\right)+2\mathrm{b}=20 \\ \Rightarrow 2\mathrm{b}=14 \\ \Rightarrow \mathrm{b}=7 \\ \therefore \mathrm{Required} \mathrm{points} \mathrm{A}=(-4, 7), \mathrm{B}=(1, 2) \end{gathered}$$