The distance of the point \((-1,1)\) from the line \(12(x+6)=5(y-2)\) is

The given line is \(12(x+6)=5(y-2)\) \(\Rightarrow 12 x+72=5 y-10\) or \(12 x-5 y+72+10=0 \Rightarrow 12 x-5 y+82=0\) The perpendicular distance from \(\left(x_1, y_1\right)\) to the line \(a x+b y+c=0\) is \(\frac{\left(\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}\right)}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\). The point \(\left(x_1, y_1\right)\) is \((-1,1)\), therefore, perpendicular distance from \((-1,1)\) to the line \(12 x-5 y+82=0\) is \(=\frac{|-12-5+82|}{\sqrt{12^2+(-5)^2}}=\frac{65}{\sqrt{144+25}}=\frac{65}{\sqrt{169}}=5\)