Let AB be the chord $4\mathrm{x}-3\mathrm{y}+5=0$ of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}-20=0.$ If C = (7, 1) then the area of triangle ABC is

$\mathrm{h}=\perp \text{ distane from }C\left(7,1\right)\text{ to this line 4x-3y+5=0}$

$\begin{array}{l}h=\frac{\left|4\left(7\right)-3\left(1\right)+5\right|}{\sqrt{16+9}}=\frac{30}{5}=6\\ \mathrm{d}=\mathrm{perpendicular} \mathrm{distance} \mathrm{from} \mathrm{centre} (1,-2) \mathrm{to} \mathrm{the} \mathrm{chord}=\frac{\left|4\left(1\right)-3\left(-2\right)+5\right|}{\sqrt{16+9}}=3\\ \mathrm{AB}=2\sqrt{{\mathrm{r}}^2-{\mathrm{d}}^2}=2\sqrt{25-9}=8\\ \text{ area }=\frac{1}{2}\left(\mathrm{AB}\right)\times \mathrm{h}=\frac{1}{2}\left(8\right)\left(6\right)=24\end{array}$