Erik Maia - Distância de Ponto à Reta

Exercícios\n1. a) A(4,6) B(1,2) d=√(1-(-1)²+\n(4-6)²) \nd=√(3²+(-1)²) \nd=√9+1 \nd=√10\n d=25\n d=5 \nd=5\n d=5\ng) A(4,4) B(5,-3) d=√(5-4)²+\n d=√(1+7)²)\n d=√1+49 \nd=√50 \nd=√50 = 5 \nd= 10\n2. A(0,7) B(8,0) d=√(8-0)²+\n d=√(7²+0)²\nd=√(64+49)\n d=√(114)\n d=√(x²+z²)\n3 III. L(1/2, 3/2) M(4,7)\n d=√(3-2)²+\n (4-1)²) d=√3²+6\n d=√(−8)²+64\n d=√18=√3 d=10\n 1\n2. C(18,13) D(√3,1/2)\n d=3-1/2) 2) a) A(a+3, 3a) B(2a+8,-5) d=10\n (10²=(2a+8)-3)²+(-5-3)²\n(10²=(a+5)²+(2+3)²\n 4+40a=50=0 \na²+4a-5=0\n A= 4=-4,1,5 a²=46. a=2 a²=1\nA=(6+2= 2 \nA= 36 \na=√4/36 a= ±46 /3 10 a²=5\n 2 1\n a= 1/6 \na=-1/2 2\n A=12-4;3,4\n a=-12. a=-12\nA=144-80 a=64 \n a= -12(3/64)\n =1.2 A=0\n a=1.2\n 1) A(3,-5) B(4,2) n=2\nAP: 2 Xp-X2=2\nYP: YP-Y2=2\nPB: X2-XP\nY2-Yp\nXp=3-2\nYp+5=2\nY-Yp=2-Yp\nXp=5-4-2Y\n3X=11\n3Y=1\nYp=11\n3\nP(11,-1)\n3-3\nD) (1,3) M(9:8) n=2\nP: 2 Yp-1=2\nYP-Yp=2\nPM 0-Xp\nXp+2/2 Yp\nXp=2\nP(1/2,-19)\n P(2) A(3,8) n=4\nAP: =4 1-3:4\nPB: X2-X4=4\nY2-Yp=2\n2-4X0=4 6:4Y=8\nY2=2\nYX0=3\nY2=Y2\n\na) P(1,-9)\nB(1/2,1/2)\n B(4,-1) P(5,2) n=-2\nAB=2 4-YA=2\n1P 5-4\n4-XA=10+x\n-YA=5\nYA=6\nA(6,5)\n 1) (5, 2) A(4, 3)\n-1-xB 3 = y0\n2 - 2 = 2\n3: 3: yB\nY0 = 2\nB(2, 2)\nY0 = 5/2\nYB = 5/3\nY0 = 5/2\n\n9) A(2, 5) B(2, 4)\n2: 2: yX\n2\n2: 2: yA\n6: 2: 4 yW\nXW = 1\nYN = 3\n\nD) A(-1, 2) N(2, 4)\nY0 = -1 + 2: 2\nY0 = 1\nY0 = -1/2\n\n10) A(5, 1) B(2, 3) C(6, 3)\nX0 = 5x + 4 = 19\nX0 =\nY0 = 5 + 3x + 4\nG(5, 4) B1 A(4, 7) B(5, 9) (10, 9)\nY = 4x + 0 × 5 + Y0 = 4 + Y0 - 3\n6(5, 2)\nI A(4, 2) B(3, 5) G(0, 0)\n0: 0 = x\nX = 0\nY0 = A\n\nT) A(6, 1) B(5, 3) (6, 3)\n\n6(-6, 7)\nY = 3\n3 = 4: 4Y\nXW = M = 3Y\nG(-11, 5, 2)\nA(3, 21) D A(4, -5) B(2, 3) (1, 1) 6(a, b) 6(3, 4)\na1 = 4x + 3 + a\n3\n2x + 3 = a6\n2 = 3 = 3 = 3\n\nb1 = 5 + 5Y + 5; 5 + 3M + 6 = Y\n2x + 2 = 4\n\nA(2, 3)\n\n1) E A(1) B(2, 3) (3, 3)\nA = 1 + 6 + 6 - 10 - 12 - 15\n3 4 5 5 1: 1\nOs pontos não são colineares\n\n2) (-4, 0) M(0, 5) N(14)\nA = 4 - 0 + 0 + 1| = 2: 4 + 5 = 0\n0, 6 = 0 = -46\nOs pontos não são colineares A(3, 10) B(3) C(10) |\nA = 0\nm = 3 m 0 3 | 3i - 3m - 15 - 10m\n\nm = 78/3\n\nA(2) B(2) C(m) A = 0\nA = 3 6 2 (other option is 2 which could also work)\nm = 0\nm = 4\nm = 0\n\nA = 12 (14 + 10 3)\nA = 2\n2 2 A = 0\n\nA = 7m + 10 = 0. A(4) A(1) B(2) C(3) A = 0\nA = 3 m 0 2 3 | (2 3) B 2 3) (m m) (m m) (m = 0\nA = 1 0 0 + 9 + 2 0\nm = 78\nA = 4(m - 2) + 12 + 10\nm = 15\nA = 6 0\nA = 4 m = 6 m = 0 A(1) B(2) C(3)\nA = 3 m 0 3 | 3i - 3m(10) - 15 - 10m\n10 1 5 10\n\nm = 78/3\n\nA(2) B(2 - 6, 3) C(m) A = 0\nA = 1 2 3 m - 2\n2 0\n\nA = 3 6 | And choose m to represent this. 2m 3m = 0\nm = 5\n\nA(3) B(3) C(3) A = 0\nA = 4 m 1 (for m = 1)\nA = 3 m 2 3\nm = 20 3\n7 3\n\nm = 14 2\n6x + 3 6\nm = 5\nx = 3 m = 1/3\n