Đáp án: $(a,b)\in\{(-2,3) , (-\dfrac12,\dfrac32)\}$
Giải thích các bước giải:
Ta có $A(1,1)\in (d)\to 1=a\cdot 1+b\to a+b=1\to b=1-a$
Lại có: $(d): y=ax+b\to ax-y+b=0$
$\to d(O,d)=\dfrac{|a\cdot 0-0+b|}{\sqrt{a^2+(-1)^2}}=\dfrac{3\sqrt{5}}{5}$
$\to \dfrac{|b|}{\sqrt{a^2+1}}=\dfrac{3\sqrt{5}}{5}$
$\to \dfrac{|1-a|}{\sqrt{a^2+1}}=\dfrac{3\sqrt{5}}{5}$
$\to (\dfrac{|1-a|}{\sqrt{a^2+1}})^2=(\dfrac{3\sqrt{5}}{5})^2$
$\to \dfrac{(1-a)^2}{a^2+1}=\dfrac95$
$\to \left(1-a\right)^2\cdot \:5=\left(a^2+1\right)\cdot \:9$
$\to a\in\{-2,-\dfrac12\}$
$\to b\in\{3,\dfrac32\}$
$\to (a,b)\in\{(-2,3) , (-\dfrac12,\dfrac32)\}$
