{"id":5395,"date":"2010-11-14T00:56:25","date_gmt":"2010-11-14T00:56:25","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=5395"},"modified":"2022-01-21T16:59:54","modified_gmt":"2022-01-21T16:59:54","slug":"escreva-uma-equacao-da-recta","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=5395","title":{"rendered":"Escreva uma equa\u00e7\u00e3o da reta&#8230;"},"content":{"rendered":"<p><ul id='GTTabs_ul_5395' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_5395' class='GTTabs_curr'><a  id=\"5395_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_5395' ><a  id=\"5395_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_5395'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Seja $(O,\\vec{i},\\vec{j})$ um referencial o.n. do plano.<\/p>\n<ol>\n<li>Escreva uma equa\u00e7\u00e3o da reta que passa no ponto $A(2,3)$ e \u00e9 perpendicular a $\\vec{u}(-1,4)$ .<\/li>\n<li>Escreva uma equa\u00e7\u00e3o da reta que passa em $B(-3,4)$ e \u00e9 perpendicular \u00e0 recta de equa\u00e7\u00e3o $2x-5y+1=0$.<\/li>\n<li>Sejam $A(2,1)$ e $B(1,5)$ dois pontos do plano.<br \/>\nDetermine uma equa\u00e7\u00e3o da mediatriz de [AB].<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_5395' onClick='GTTabs_show(1,5395)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_5395'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>O vetor $\\vec{r}(4,1)$ , perpendicular ao vetor $\\vec{u}(-1,4)$ , \u00e9 um vetor diretor da reta pedida.<br \/>\nPortanto, o declive da reta pedida \u00e9 $m=\\frac{1}{4}$, sendo a sua equa\u00e7\u00e3o reduzida do tipo $y=\\frac{1}{4}x+b$.<br \/>\nComo o ponto A \u00e9 um ponto dessa reta, as suas coordenadas t\u00eam de verificar a equa\u00e7\u00e3o anterior. Assim, temos $3=\\frac{1}{4}\\times 2+b\\Leftrightarrow b=\\frac{5}{2}$.<br \/>\nLogo, $y=\\frac{1}{4}x+\\frac{5}{2}$ \u00e9 a equa\u00e7\u00e3o reduzida da reta pedida.<br \/>\n\u00ad<\/li>\n<li>Como $2x-5y+1=0\\Leftrightarrow y=\\frac{2}{5}x+\\frac{1}{5}$, a reta dada tem declive $m=\\frac{2}{5}$.<br \/>\nComo a reta pedida \u00e9 perpendicular a esta reta, o seu declive ser\u00e1 $m&#8217;=-\\frac{1}{m}=-\\frac{5}{2}$ e a sua equa\u00e7\u00e3o reduzida da forma $y=-\\frac{5}{2}x+b$.<br \/>\nComo o ponto B \u00e9 um ponto dessa reta, as suas coordenadas t\u00eam de verificar a equa\u00e7\u00e3o anterior. Assim, temos $4=-\\frac{5}{2}\\times (-3)+b\\Leftrightarrow b=-\\frac{7}{2}$.<br \/>\nLogo, $y=-\\frac{5}{2}x-\\frac{7}{2}$ \u00e9 a equa\u00e7\u00e3o reduzida da reta pedida.<br \/>\n\u00ad<\/li>\n<li>O vetor $\\vec{r}=(4,1)$ , perpendicular ao vetor $\\overrightarrow{AB}=(-1,4)$, \u00e9 um vetor diretor da mediatriz de [AB].<br \/>\nPor outro lado, essa reta passa no ponto m\u00e9dio de [AB]: $M(\\frac{2+1}{2},\\frac{1+5}{2})=(\\frac{3}{2},3)$.<br \/>\nLogo, a equa\u00e7\u00e3o da mediatriz de [AB] \u00e9 do tipo $y=\\frac{1}{4}x+b$. Dado que M \u00e9 um ponto dessa reta, temos $3=\\frac{1}{4}\\times \\frac{3}{2}+b\\Leftrightarrow b=\\frac{21}{8}$.<br \/>\nLogo, $y=\\frac{1}{4}x+\\frac{21}{8}$ \u00e9 a equa\u00e7\u00e3o reduzida da mediatriz de [AB].<br \/>\n\u00ad<\/li>\n<\/ol>\n<blockquote>\n<p><strong>Uma reta r de equa\u00e7\u00e3o $y=mx+b$ admite como vetor diretor o vetor $\\vec{r}(1,m)$ \u00a0 e como vetor normal $\\vec{n}(-m,1)$ .<\/strong><\/p>\n<p><strong>Se $m\\ne 0$, ent\u00e3o $-\\frac{1}{m}$ \u00e9 o declive das retas perpendiculares a r.<\/strong><\/p>\n<p><strong>Isto \u00e9, retas perpendiculares n\u00e3o paralelas aos eixos coordenados possuem declives tais que: o declive de uma das retas \u00e9 sim\u00e9trico do inverso do declive da outra recta, ou seja, $m&#8217;=-\\frac{1}{m}$.<\/strong><\/p>\n<\/blockquote>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_5395' onClick='GTTabs_show(0,5395)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Seja $(O,\\vec{i},\\vec{j})$ um referencial o.n. do plano. Escreva uma equa\u00e7\u00e3o da reta que passa no ponto $A(2,3)$ e \u00e9 perpendicular a $\\vec{u}(-1,4)$ . Escreva uma equa\u00e7\u00e3o da reta que passa em&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20826,"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-5395","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":2197,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/11\/11V1Pag180-21_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/5395","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=5395"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/5395\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20826"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5395"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5395"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5395"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=5395"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}