A' cleachdadh d' eòlais
Tha na co-chomharran \(P( - 3, - 4)\), \(Q( - 3,4)\) agus \(R(5,12)\) aig an triantan PQR.
Question
a) Obraich a-mach co-aontar na loidhne-meadhain \(MR\)
a) \(M\left( {\frac{{ - 3 + ( - 3)}}{2},\frac{{4 + ( - 4)}}{2}} \right) = M( - 3,0)\)
Airson \(M( - 3,0)\) agus \(R(5,12)\)
\({m_{MR}} = \frac{{12 - 0}}{{5 - ( - 3)}} = \frac{{12}}{8} = \frac{3}{2}\)
'S e caisead na loidhne-meadhain \(\frac{3}{2}\)
'S e a' phuing air an loidhne-meadhain \(( - 3,0)\)
Mar sin 's e an co-aontar \(y - 0 = \frac{3}{2}\left( {x - ( - 3)} \right)\)
\(2y = 3(x + 3)\)
\(2y=3x+9\)
\(2y - 3x = 9\)
\(3x-2y+9=0\)
Question
b) Obraich a-mach an loidhne-àirde \(NQ\)
b) Airson \(P( - 3, - 4)\) agus \(R(5,12)\):
\({m_{PR}} = \frac{{12 - ( - 4)}}{{5 - ( - 3)}} = \frac{{16}}{8} = 2\)
\(\Rightarrow {m_ \bot } = - \frac{1}{2}\,\,\,\,\,\,\,\,\,\,(bhon\,a\,tha\,\, m_{1}\times m_{2} = -1)\)
'S e caisead na loidhne-àirde \(- \frac{1}{2}\)
'S e a' phuing air an loidhne-àirde \(Q( - 3,4)\)
Mar sin 's e an co-aontar \(y - 4 = - \frac{1}{2}(x - ( - 3))\)
\(2y - 8 = - (x + 3)\)
\(2y - 8 = - x - 3\)
\(2y + x = 5\)
Question
c) Tha loidhne-meadhain \(MR\) agus loidhne-àirde \(NQ\) a' trasnadh aig puing \(S\). Obraich a-mach na co-chomharran aig \(S\).
c) Gus a' phuing-trasnaidh obrachadh a-mach, fuasgail:
\(2y + x = 5\)
\(2y - 3x = 9\)
Thoir-air-falbh:
\(4x = - 4\)
\(\Rightarrow x = - 1\)
Ionadaich \(x = - 1\) do \(2y + x = 5\)
\(2y-1=5\)
\(2y = 6\)
\(y = 3\)
'S e a' phuing-trasnaidh \(S( - 1,3)\)
Question
d) Tha a' phuing \(T( 2, 9)\) air \(QR\). Seall gu bheil \(ST\) co-shìnte ri \(PR\).
d) Bho b) \({m_{PR}} = 2\)
Airson \(S( - 1,3)\) agus \(T(2,9)\)
\({m_{ST}} = \frac{{9 - 3}}{{2 - ( - 1)}} = \frac{6}{3} = 2\)
Mar sin \({m_{ST}} = {m_{PR}} = 2\)
Mar sin tha \(ST\) co-shìnte ri \(PR\).