{"id":326,"date":"2020-03-27T21:46:47","date_gmt":"2020-03-28T00:46:47","guid":{"rendered":"http:\/\/cinoto.com.br\/matematica\/?p=326"},"modified":"2020-03-27T21:46:47","modified_gmt":"2020-03-28T00:46:47","slug":"determine-a-abscissa-xb-de-um-dos-vertices-do-triangulo-abc","status":"publish","type":"post","link":"http:\/\/cinoto.com.br\/matematica\/determine-a-abscissa-xb-de-um-dos-vertices-do-triangulo-abc\/","title":{"rendered":"2) Determine a abscissa Xb de um dos v\u00e9rtices do triangulo ABC, cuja \u00e1rea vale 15 e os v\u00e9rtices s\u00e3o: A(8, 3), B(Xb, 7) e C(2, 1)."},"content":{"rendered":"\n\n\n\n\n

Resolu\u00e7\u00e3o:<\/h2>\n\n\n\n

Nesse exerc\u00edcio, podemos usar a f\u00f3rmula que nos d\u00e1 a \u00e1rea de um tri\u00e2ngulo segundo suas coordenadas cartesianas. Sejam os v\u00e9rtices do tri\u00e2ngulo ABC os pontos A = (Xa<\/sub>, Ya<\/sub>), B = (Xb<\/sub>, Yb<\/sub>) e C = (Xc<\/sub>, Yc<\/sub>). Sua \u00e1rea \u00e9 igual a:
= (1\/2).|Xa<\/sub>\u00a0 Ya<\/sub>\u00a0 1|
\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 |Xb<\/sub>\u00a0 Yb<\/sub>\u00a0 1|
\u00a0\u00a0 \u00a0\u00a0 \u00a0\u00a0 \u00a0\u00a0\u00a0 |Xc<\/sub>\u00a0 Yc<\/sub>\u00a0 1|<\/p>\n\n\n\n

\u00c9 igual a metade do m\u00f3dulo desse determinante. Ent\u00e3o como a \u00e1rea vale 15 e s\u00f3 falta uma coordenada, vamos substituir o resto:
\u00e1rea = (1\/2).|Xa<\/sub>\u00a0 Ya<\/sub>\u00a0 1|
\u00a0\u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0 |Xb<\/sub>\u00a0 Yb<\/sub>\u00a0 1|
\u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0 \u00a0 \u00a0\u00a0 |Xc<\/sub>\u00a0 Yc<\/sub>\u00a0 1|<\/p>\n\n\n\n

15 = (1\/2).|8   3  1|
                   |Xb<\/sub>  7  1|
                   |2   1  1|<\/p>\n\n\n\n

15 = (1\/2).(56 + 6 + Xb<\/sub> – 14 – 8 – 3Xb<\/sub>)
15 = (1\/2).(40 – 2Xb<\/sub>)
15 = 20 – Xb<\/sub>
Xb<\/sub> = 5<\/p>\n\n\n\n

Resposta: A abcissa Xb<\/sub>\u00a0vale 5.<\/p>\n","protected":false},"excerpt":{"rendered":"","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12],"tags":[28],"class_list":["post-326","post","type-post","status-publish","format-standard","hentry","category-geometria-analitica","tag-facil"],"_links":{"self":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts\/326","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/comments?post=326"}],"version-history":[{"count":1,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts\/326\/revisions"}],"predecessor-version":[{"id":327,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/posts\/326\/revisions\/327"}],"wp:attachment":[{"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/media?parent=326"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/categories?post=326"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/cinoto.com.br\/matematica\/wp-json\/wp\/v2\/tags?post=326"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}