Question

VA ROG FRUMOS! TOT PUNCTUL 1 ROMAN

Answer 1

$1. \A(-4, 7),B(-1,3)\ => AB = \sqrt{(-1-4)^2+(3-7)^2}=\sqrt{(-5)^2+(-4)^2}\\\\AB=\sqrt{25+16}=\sqrt{41}\\\\2.\ A(-a,5), B(0,-3)\ => AB = \sqrt{(0+a)^2+(-3-5)^2}= \sqrt{(a^2+(-8)^2}\\\\10=\sqrt{a^2+64}\ => 100 = a^2 +64\ => a^2 = 36 => a_1= 6, a_2=-6\\\\3.\ x_M= (2-4) : 2 = -1\\\\y_M= (9+5) : 2 = 14 : 2 =7 \\\\4.\ x_M=(-3+x_B) : 2\ => -5= (-3+x_B) : 2\ => -10 = -3+x_B\\\\x_B= -7\\\\y_m=(8+y_B) : 2 = 12 => 12= (8+y_B) : 2\ => 24 = 8+y_B\\\\y_B= 16$$5.\ M(-5,6), N(3,0), P(0,-4)\\\\MN=\sqrt{(3+5)^2+6^2}= \sqrt{8^2+36}=\sqrt{100} = 10\\\\NP=\sqrt{3^2+(-4)^2}= \sqrt{9+16}=\sqrt{25} = 5\\\\MP=\sqrt{5^2+(-4-6)^2}= \sqrt{25+(-10)^2}=\sqrt{25+100} = \sqrt{125}=5\sqrt{5}\\\\P= MN + NP + MP = 10 + 5 + 5\sqrt{5} = 15 + 5\sqrt{5}\\\\6.\ A(a,2), B(3, a-1), AB= 2\sqrt{2}\\AB=\sqrt{(3-a)^2+(a-1-2)^2}=\sqrt{9-6a+a^2+a^2-6a+9}\\\\AB=\sqrt{2(a-3)^2}=2\sqrt{2}\\2(a-3)^2= 8\\\\(a-3)^2=4\\\\a_1-3=4\ => a_1=4+3=7\\a_2-3=-4\ => a_2=-4+3=-1$