{"id":5534,"date":"2010-11-15T02:51:14","date_gmt":"2010-11-15T02:51:14","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=5534"},"modified":"2022-01-12T12:14:02","modified_gmt":"2022-01-12T12:14:02","slug":"rascunho-29","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=5534","title":{"rendered":"Equa\u00e7\u00e3o de uma reta que passa em A e \u00e9 perpendicular a r"},"content":{"rendered":"<p><ul id='GTTabs_ul_5534' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_5534' class='GTTabs_curr'><a  id=\"5534_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_5534' ><a  id=\"5534_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<li id='GTTabs_li_2_5534' ><a  id=\"5534_2\" onMouseOver=\"GTTabsShowLinks('E se&#8230;'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>E se&#8230;<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_5534'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Considere, num referencial o. n. $(O,\\vec{i},\\vec{j})$, a reta r de equa\u00e7\u00e3o $(x,y)=(3,2)+k(-3,-1),k\\in \\mathbb{R}$ e o ponto $A(-1,4)$.<\/p>\n<ol>\n<li>Determine a equa\u00e7\u00e3o reduzida da reta s, perpendicular a r e que passa em A.<\/li>\n<li>Desenhe um quadrado de v\u00e9rtice A, com um lado sobre a reta s e outro sobre a reta r, e determine, analiticamente, as coordenadas dos v\u00e9rtices do quadrado que construiu.<\/li>\n<\/ol>\n<p>\u00a0<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_5534' onClick='GTTabs_show(1,5534)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_5534'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p>Execute a anima\u00e7\u00e3o seguinte:<\/p>\n<p style=\"text-align: center;\"><script src=\"https:\/\/cdn.geogebra.org\/apps\/deployggb.js\"><\/script>\r\n<div id=\"ggbApplet\" style=\"margin: 0 auto;\"><\/div>\r\n<script>\r\nvar parameters = {\r\n\"id\": \"ggbApplet\",\r\n\"width\":560,\r\n\"height\":380,\r\n\"showMenuBar\":false,\r\n\"showAlgebraInput\":false,\r\n\"showToolBar\":false,\r\n\"customToolBar\":\"0 39 | 1 501 67 , 5 19 , 72 | 2 15 45 , 18 65 , 7 37 | 4 3 8 9 , 13 44 , 58 , 47 | 16 51 64 , 70 | 10 34 53 11 , 24  20 22 , 21 23 | 55 56 57 , 12 | 36 46 , 38 49  50 , 71 | 30 29 54 32 31 33 | 17 26 62 73 , 14 68 | 25 52 60 61 | 40 41 42 , 27 28 35 , 6\",\r\n\"showToolBarHelp\":false,\r\n\"showResetIcon\":true,\r\n\"enableLabelDrags\":false,\r\n\"enableShiftDragZoom\":false,\r\n\"enableRightClick\":false,\r\n\"errorDialogsActive\":false,\r\n\"useBrowserForJS\":false,\r\n\"preventFocus\":false,\r\n\"showFullscreenButton\":true,\r\n\"language\":\"pt\",\r\n\/\/ use this instead of ggbBase64 to load a material from GeoGebraTube\r\n\/\/ 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is3D=is 3D applet using 3D view, AV=Algebra View, SV=Spreadsheet View, CV=CAS View, EV2=Graphics View 2, CP=Construction Protocol, PC=Probability Calculator, DA=Data Analysis, FI=Function Inspector, PV=Python, macro=Macro View\r\nvar views = {'is3D': 0,'AV': 0,'SV': 0,'CV': 0,'EV2': 0,'CP': 1,'PC': 0,'DA': 0,'FI': 0,'PV': 0,'macro': 0};\r\nvar applet = new GGBApplet(parameters, '5.0', views);\r\nwindow.onload = function() {applet.inject('ggbApplet')};\r\n<\/script><\/p>\n<p style=\"text-align: left;\">Vamos resolver o exerc\u00edcio com base no procedimento apresentado.<\/p>\n<ol>\n<li>Um vector director da reta r \u00e9 $\\vec{r}(-3,-1)$.\n<p>Designado por $P(x,y)$ um ponto gen\u00e9rico da reta s, ser\u00e1 $\\overrightarrow{AP}.\\vec{r}=0$.<\/p>\n<p>Logo, uma condi\u00e7\u00e3o que define a reta s \u00e9 \\[\\begin{array}{*{35}{l}}<br \/>\n\\overrightarrow{AP}.\\vec{r}=0 &amp; \\Leftrightarrow\u00a0 &amp; (x+1,y-4).(-3,-1)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; -3x-3-y+4=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; y=-3x+1\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nPortanto, $y=-3x+1$ \u00e9 a equa\u00e7\u00e3o reduzida da reta s.<\/p>\n<\/li>\n<li>Um\u00a0dos v\u00e9rtices do quadrado \u00e9 j\u00e1 conhecido: $A(-1,4)$.\n<p>Determinemos as coordenadas de um segundo v\u00e9rtice do quadrado: ponto Q, ponto de intersec\u00e7\u00e3o das retas r e s.<\/p>\n<p>Como $(x,y)=(3,2)+k(-3,-1),k\\in \\mathbb{R}$ \u00e9 uma equa\u00e7\u00e3o vetorial da reta r, ent\u00e3o ${{m}_{r}}=\\frac{-1}{-3}=\\frac{1}{3}$ (Porqu\u00ea?) e como o ponto de coordenadas $(0,1)$ (obt\u00e9m-se da equa\u00e7\u00e3o vetorial, para $k=1$) \u00e9 um ponto dessa reta, a sua equa\u00e7\u00e3o reduzida \u00e9 $y=\\frac{1}{3}x+1$.<\/p>\n<p>Assim, \\[\\begin{matrix}<br \/>\n\\left\\{ \\begin{array}{*{35}{l}}<br \/>\ny=-3x+1\u00a0 \\\\<br \/>\ny=\\frac{1}{3}x+1\u00a0 \\\\<br \/>\n\\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}}<br \/>\ny=-3x+1\u00a0 \\\\<br \/>\n-3x+1=\\frac{1}{3}x+1\u00a0 \\\\<br \/>\n\\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}}<br \/>\nx=0\u00a0 \\\\<br \/>\ny=1\u00a0 \\\\<br \/>\n\\end{array} \\right.\u00a0 \\\\<br \/>\n\\end{matrix}\\]<\/p>\n<p>Logo, o ponto $Q(0,1)$ \u00e9 um segundo v\u00e9rtice do quadrado.<\/p>\n<p>Dado que $\\overline{QB}=\\overline{QA}=\\sqrt{{{(3-0)}^{2}}+{{(2-1)}^{2}}}=\\sqrt{{{(-1-0)}^{2}}+{{(4-1)}^{2}}}=\\sqrt{10}$, ent\u00e3o um terceiro v\u00e9rtice do quadrado \u00e9 (pode ser)\u00a0$B(3,2)$. (Ver a aba &#8220;E se&#8230;&#8221;)<\/p>\n<p>O quarto v\u00e9rtice, ponto T,\u00a0pode ser determinado tendo em considera\u00e7\u00e3o que $T=A+\\overrightarrow{QB}$.<br \/>\nDonde, $T=A+\\overrightarrow{QB}=(-1,4)+(3,1)=(2,5)$.<\/p>\n<p><span style=\"text-decoration: underline;\">Mas, o problema tem outra solu\u00e7\u00e3o<\/span>:<\/p>\n<p>Outra solu\u00e7\u00e3o \u00e9 o quadrado [AQB&#8217;T&#8217;], sendo $B&#8217;=B-\\overrightarrow{QB}=(0,1)-(3,1)=(-3,0)$ e $T&#8217;=A-\\overrightarrow{QB}=(-1,4)-(3,1)=(-4,3)$. (Ver a aba &#8220;E se&#8230;&#8221;)<\/p>\n<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_5534' onClick='GTTabs_show(0,5534)'>&lt;&lt; Enunciado<\/a><\/span><span class='GTTabs_nav_next'><a href='#GTTabs_ul_5534' onClick='GTTabs_show(2,5534)'>E se&#8230; &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_2_5534'>\n<span class='GTTabs_titles'><b>E se&#8230;<\/b><\/span><\/p>\n<p>Reparou-se que $\\overline{QB}=\\overline{QA}=\\sqrt{{{(3-0)}^{2}}+{{(2-1)}^{2}}}=\\sqrt{{{(-1-0)}^{2}}+{{(4-1)}^{2}}}=\\sqrt{10}$, pelo que um terceiro v\u00e9rtice do quadrado \u00e9 (pode ser)\u00a0$B(3,2)$.<\/p>\n<p>E se n\u00e3o se tivesse reparado nesse pormenor?<\/p>\n<p style=\"text-align: center;\"><div id=\"ggbApplet2\" style=\"margin: 0 auto;\"><\/div>\r\n<script>\r\nvar parameters = {\r\n\"id\": 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is3D=is 3D applet using 3D view, AV=Algebra View, SV=Spreadsheet View, CV=CAS View, EV2=Graphics View 2, CP=Construction Protocol, PC=Probability Calculator, DA=Data Analysis, FI=Function Inspector, PV=Python, macro=Macro View\r\nvar views = {'is3D': 0,'AV': 0,'SV': 0,'CV': 0,'EV2': 0,'CP': 1,'PC': 0,'DA': 0,'FI': 0,'PV': 0,'macro': 0};\r\nvar applet2 = new GGBApplet(parameters, '5.0', views);\r\nwindow.onload = function() {applet.inject('ggbApplet'); applet2.inject('ggbApplet2')};\r\n<\/script><\/p>\n<p><strong>M\u00e9todo 1<\/strong>:<\/p>\n<p>Os vetores perpendiculares ao vetor $\\overrightarrow{QA}=(-1,3)$, com norma igual \u00e0 deste vetor, s\u00e3o os vetores $\\overrightarrow{u}=(3,1)$ e $\\overrightarrow{v}=(-3,-1)$ (vetores sim\u00e9tricos).<\/p>\n<p>Os pontos B e T podem ser assim obtidos: $B=Q+\\overrightarrow{u}=(0,1)+(3,1)=(3,2)$ e $T=A+\\overrightarrow{u}=(-1,4)+(3,1)=(2,5)$.<\/p>\n<p>Os pontos B&#8217; e T&#8217; podem ser assim obtidos: $B&#8217;=Q+\\overrightarrow{v}=(0,1)+(-3,-1)=(-3,0)$ e $T&#8217;=A+\\overrightarrow{v}=(-1,4)+(-3,-1)=(-4,3)$.<\/p>\n<p><strong>M\u00e9todo 2<\/strong>:<\/p>\n<p>Os pontos B e B\u00b4 pertencem \u00e0 reta r, mas ter\u00e1 de ser $\\overline{QA}=\\overline{QB}=\\overline{QB&#8217;}$ (Porqu\u00ea?).<\/p>\n<p>Logo, os pontos B e B&#8217; pertencem simultaneamente \u00e0 reta r e \u00e0 circunfer\u00eancia com centro em Q e raio [QA].<\/p>\n<p>Assim, vem: \\[\\begin{matrix}<br \/>\n\\left\\{ \\begin{array}{*{35}{l}}<br \/>\n{{x}^{2}}+{{(y-1)}^{2}}=10\u00a0 \\\\<br \/>\ny=\\frac{1}{3}x+1\u00a0 \\\\<br \/>\n\\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}}<br \/>\n{{x}^{2}}+{{(\\frac{1}{3}x)}^{2}}=10\u00a0 \\\\<br \/>\ny=\\frac{1}{3}x+1\u00a0 \\\\<br \/>\n\\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}}<br \/>\n\\frac{10}{9}{{x}^{2}}=10\u00a0 \\\\<br \/>\ny=\\frac{1}{3}x+1\u00a0 \\\\<br \/>\n\\end{array} \\right. &amp; \\Leftrightarrow\u00a0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}}<br \/>\nx=-3\\vee x=3\u00a0 \\\\<br \/>\ny=\\frac{1}{3}x+1\u00a0 \\\\<br \/>\n\\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}}<br \/>\nx=-3\u00a0 \\\\<br \/>\ny=0\u00a0 \\\\<br \/>\n\\end{array} \\right.\\vee \\left\\{ \\begin{array}{*{35}{l}}<br \/>\nx=3\u00a0 \\\\<br \/>\ny=2\u00a0 \\\\<br \/>\n\\end{array} \\right. &amp; {}\u00a0 \\\\<br \/>\n\\end{matrix}\\]<\/p>\n<p>Portanto, $B(3,2)$ e $B'(-3,0)$, donde $T=A+\\overrightarrow{QB}=(-1,4)+(3,1)=(2,5)$ e $T&#8217;=A-\\overrightarrow{QB}=(-1,4)-(3,1)=(-4,3)$.<\/p><\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_5534' onClick='GTTabs_show(1,5534)'>&lt;&lt; Resolu\u00e7\u00e3o<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Considere, num referencial o. n. $(O,\\vec{i},\\vec{j})$, a reta r de equa\u00e7\u00e3o $(x,y)=(3,2)+k(-3,-1),k\\in \\mathbb{R}$ e o ponto $A(-1,4)$. Determine a equa\u00e7\u00e3o reduzida da reta s, perpendicular a r e que passa em&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19178,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,110],"tags":[422,67,111],"series":[],"class_list":["post-5534","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-geometria-analitica","tag-11-o-ano","tag-geometria","tag-vectores"],"views":2012,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat69.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/5534","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=5534"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/5534\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19178"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5534"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5534"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5534"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=5534"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}