The perpendicular distance from the point $(-2,-3)$ to the line $12x+16y+12=0$ is:

1. Use $d=\left|\dfrac{Ah+Bk+C}{\sqrt{A^2+B^2}}\right|$ with $A=12$, $B=16$, $C=12$. 2. Numerator: $12(-2)+16(-3)+12=-24-48+12=-60$. 3. Denominator: $\sqrt{12^2+16^2}=\sqrt{144+256}=\sqrt{400}=20$. 4. $d=\left|\dfrac{-60}{20}\right|=3$ units. 5. Forgetting the modulus or dropping the constant term $12$ produces the other values. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 8: Coordinate Geometry I_