Co-aontar loidhneachan-àirde is letheadairean ceart-cheàrnach
Cuimhnich air an eòlas a th' agad mu bhith ag obrachadh a-mach co-aontar loidhne agus cleachd e airson na loidhne bhriste anns na triantain seo.
Cuimhnich nuair a bhios tu ag obrachadh a-mach co-aontar loidhne gum feum thu puing \((a,b)\) agus an caisead \(m\).
Co-aontar loidhne-meadhain
Question
Obraich a-mach co-aontar na loidhne-meadhain tro A ann an triantan ABC.
Tha an loidhne-meadhain a' dol tron phuing-meadhain aig BC.
\(= \left( {\frac{{6 + 4}}{2},\frac{{3 + 11}}{2}} \right)\)
\(= (5,7)\)
Faodaidh sinn a-nis a' phuing-meadhain agus co-chomharran na puing A a chleachdadh gus an caisead \(m\) obrachadh a-mach.
\(m = \frac{{7 - 2}}{{5 - 1}} = \frac{5}{4}\)
Cleachd \((1,2)\) no \((5,7)\) agus \(m\) gus an co-aontar obrachadh a-mach.
\(y - b = m(x - a)\)
\(y - 2 = \frac{5}{4}(x - 1)\)
\(4y - 8 = 5x - 5\)
\(4y = 5x + 3\)
\(5x-4y+3=0\)
Co-aontar loidhne-àirde
Question
Obraich a-mach co-aontar na loidhne-àirde tro B ann an triantan ABC.
Tha an loidhne-àirde ceart-cheàrnach ri AC:
\({m_{AC}} = \frac{{11 - 2}}{{4 - 1}} = \frac{9}{3} = 3\)
Tha caiseadan ceart-cheàrnach co-cheangailte le \({m_1}\times {m_2} = - 1\)
\({m_{perp}} = - \frac{1}{3}\,\,\,\,\,\,\,\,\,\,\,\,\, (bhon\,a\,tha - \frac{1}{3} \times 3 = - 1)\)
Tha seo ag obrachadh a-mach caisead na loidhne a tha a dhìth.
Cleachd B\((6,3)\) agus \(m\).
\(y - b = m(x - a)\)
\(y - 3 = - \frac{1}{3}(x - 6)\)
\(3y - 9 = - x + 6\)
\(3y + x = 15\)
\(x+3y-15=0\)
Co-aontaran Letheadair Ceart-cheàrnach
Question
Obraich a-mach co-aontar an letheadair cheart-cheàrnaich aig AC ann an triantan ABC.
Tha an letheadair ceart-cheàrnach a' dol tro phuing-meadhain AC. Tha seo a' suidheachadh na puing air an loidhne a tha a dhìth.
Tha puing-meadhain AC \(= \left( {\frac{{1 + 4}}{2},\frac{{2 + 11}}{2}} \right) = \left( {\frac{5}{2},\frac{{13}}{2}} \right)\)
Tha an letheadair ceart-cheàrnach gu ceart-cheàrnach ri AC.
\({m_{AC}} = \frac{{11 - 2}}{{4 - 1}} = \frac{9}{3} = 3\)
Tha caiseadan ceart-cheàrnach air an ceangal le \({m_1}\times{m_2} = - 1\). Tha seo ag obrachadh a-mach caisead na loidhne a tha a dhìth.
\({m_{perp}} = - \frac{1}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,(bhon\,a\,tha\,- \frac{1}{3} \times 3 = - 1)\)
Cleachd \(\left( {\frac{5}{2},\frac{{13}}{2}} \right)\) agus \(m\).
\(y - b = m(x - a)\)
\(y - \frac{{13}}{2} = - \frac{1}{3}\left( {x - \frac{5}{2}} \right)\)
\(6y - 6 \times \frac{{13}}{2} = 6 \times \left( { - \frac{1}{3}\left( {x - \frac{5}{2}} \right)} \right)\)
\(6y - 39 = - 2\left( {x - \frac{5}{2}} \right)\)
\(6y - 39 = - 2x + 5\)
\(6y = - 2x + 44\)
\(3y + x = 22\)
\(x + 3y - 22 = 0\)
