In \( \Large \triangle ABC \) the vertices are \( \Large A \left(-3,\ 0\right),\ B \left(4,\ -1\right)\ and\ C \left(5,\ 2\right) \)



Distance between B and C

\( \Large BC=\sqrt{ \left(5-4\right)^{2}+ \left(2+1\right)^{2} } = \sqrt{1+9} = \sqrt{10} \)

Area of \( \Large \triangle ABC \)

= \( \Large \frac{1}{2} \left[ x_{1} \left( y_{2} - y_{3} \right) + x_{2} \left( y_{3} - y_{1} \right) + x_{3} \left( y_{1} - y_{2} \right) \right] \)

= \( \Large \frac{1}{2}\left[ -3 \left(-1-2\right)+4 \left(2-0\right)+5 \left(0+1\right) \right] \)

= \( \Large \frac{1}{2}\left[ 9+8+5 \right]=11 \)

As we know area of \( \Large \triangle ABC = \frac{1}{2} \times BC \times AL \)

=> \( \Large 11 = \frac{1}{2} \times \sqrt{10} \times AL \)

=> \( \Large AL = \frac{2 \times 11}{\sqrt{10}}=\frac{22}{\sqrt{10}} \)