Răspuns:
a ∈ {-2; 7}
Rezolvare:
Fie A(2a; a-2) și B(5; 3-a), a ∈ R, AB = 9√2
[tex]AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\[/tex]
[tex]\sqrt{(5-2a)^2+(3-a-a+2)^2}=9\sqrt2\\[/tex]
[tex]\sqrt{(5-2a)^2+(5-2a)^2}=9\sqrt2\\[/tex]
[tex]\sqrt{2(25-20a+4a^2)}=9\sqrt2\\[/tex]
[tex]\sqrt{50-40a+8a^2}=9\sqrt2\\[/tex]
[tex]50-40a+8a^2=81\times2\:\:|:2\\[/tex]
[tex]25-20a+4a^2=81\\[/tex]
[tex](5-2a)^2=81\\[/tex]
[tex]5-2a=\pm9\\[/tex]
[tex]5-2a_1=+9\\[/tex]
[tex]-2a_1=4\\[/tex]
[tex]\bf a_1=-2\\[/tex]
[tex]5-2a_2=-9\\[/tex]
[tex]-2a_2=-14\\[/tex]
[tex]\bf a_2=7\\[/tex]
[tex]\implies\bf\boxed{\red{a\in\{-2;\:7\}}}\\[/tex]
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