a,
$d'// d\to d': 2x-3y+c=0$ ($c\ne 4$)
Thay $x=4; y=-1$ vào $d'$:
$2.4+3+c=0$
$\to c=-11$ (TM)
$\to d': 2x-3y-11=0$
Vậy $d(K;d')=\dfrac{|2.3+3.2-11|}{\sqrt{2^2+3^2}}=\dfrac{\sqrt{13}}{13}$
b,
$d'\bot d\to d': 3x+2y+c=0$
Thay $x=4; y=-1$ ta có:
$3.4-2+c=0$
$\to c=-10$
$\to d': 3x+2y-10=0$
Vậy $d(K;d')=\dfrac{|3.3-2.2-10|}{\sqrt{3^2+2^2}}=\dfrac{5\sqrt{13}}{\sqrt{13}}$