Vì M nằm trên (d) nên:
3a - 2b - 1 = 0 <=> 3a = 2b+1 <=> [tex]b=\frac{3a-1}{2}[/tex]
Ta có:
[tex]a^2+b^2=a^2+(\frac{3a-1}{2})^2=\frac{13a^2-6a+1}{4}=\frac{13(a-\frac{3}{13})^2+\frac{4}{13}}{4}\geq \frac{1}{13}[/tex]
Dấu "=" xảy ra <=> a = 3/13 => b = -2/13
Vậy M (3/13; -2/13)