Question

determinati a apartine R stiind ca lungimea segmentului determinat de punctele A(-1,2) si B(4-a,4+a) este egala cu 5

Answer 1

|AB| =5 <=> rad( (4-a+1)^2 + (4+a-2)^2 ) =5
<=> rad( (5-a)^2 + (2+a)^2 ) =5 / ()^2
<=> (5-a)^2 + (2+a)^2 = 25
<=> 25-10a+a^2 + 4 + 4a +a^2 = 25
<=> 2a^2 - 6a + 4 =0 / :2
<=> a^2 -3a + 2 =0
d = 9 -8 = 1
a1 = 1 ; a2 = 2 ;
a € {1,2}

Answer 2

$A(-1,2);\quad B(4-a, 4+a); \\ \\ |AB| = \sqrt{(x_B-x_A)^2+(y_B-y_A)^2} \Rightarrow \\ \Rightarrow |AB| = \sqrt{(4-a-(-1))^2+(4+a-2)^2} \Rightarrow \\ \Rightarrow |AB| = \sqrt{(4-a+1)^2+(2+a)^2} \Rightarrow \\ \Rightarrow |AB| = \sqrt{(5-a)^2+(2+a)^2} \Rightarrow \\ \Rightarrow |AB| = \sqrt{25-10a+a^2+4+4a+a^2} \Rightarrow \\ \Rightarrow |AB| = \sqrt{2a^2-6a+29} \\ \\ |AB| = 5 \Rightarrow \sqrt{2a^2-6a+29} =5 \Big|^2 \Rightarrow 2a^2-6a+29 = 25 \Rightarrow$

[tex]\Rightarrow 2a^2-6a+29-25 = 0 \Rightarrow 2a^2-6a+4 = 0 \Rightarrow a^2-3a+2 = 0 \Rightarrow \\ \\ \Rightarrow a^2-a-2a+2 = 0 \Rightarrow a(a-1)-2(a-1) = 0 \Rightarrow (a-1)(a-2) = 0 \\ \\ \Rightarrow \left| \begin{array}{c} a_1 = 1 \\ a_2=2 \end{array} \right \Rightarrow \boxed{a\in\Big\{1;2\Big\}}[/tex]