The length of the perpendicular from 1 , - 2 to the line 12 x + 5 y + 63 = 0 is

As we know length of the perpendicular from $\left(x_1,y_1,z_1\right)$ to the line $ax+by+c=0$ is $d=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

Length of the perpendicular from$\left(1,-2\right)$ to the line $12x+5y+63=0$ is $d=\left|\frac{12\left(1\right)+5\left(-2\right)+63}{\sqrt{{\left(12\right)}^2+5^2}}\right|$

$\Rightarrow d=\left|\frac{12-10+63}{\sqrt{144+25}}\right|=\left|\frac{65}{13}\right|=5$ units.