An tansaint
Bhon as e loidhne dhìreach a th' ann an tansaint, tha an co-aontar air a shealltainn san riochd \(y - b = m(x - a)\). Feumaidh tu puing agus an caisead gus an co-aontar obrachadh a-mach.
Mar as trice, bidh a' phuing air a toirt dhut – far a bheil an tansaint a' coinneachadh ris a' chearcall.
Gus an caisead obrachadh a-mach, cuimhnich gu bheil an tansaint ceart-cheàrnach ris an radius bhon phuing aig a bheil e a' coinneachadh ris a' chearcall.
Obraich a-mach caisead an radius (CP) aig a' phuing far a bheil an tansaint a' coinneachadh a' chearcaill. An uair sin cleachd an co-aontar \({m_{CP}} \times {m_{tgt}} = - 1\) agus obraich a-mach caisead an tansaint.
Eisimpleir
Obraich a-mach an tansaint ris a' chearcall \({x^2} + {y^2} - 2x - 2y - 23 = 0\) aig a' phuing \(P(5, - 2)\) a tha na laighe air a' chearcall.
'S e meadhan a' chearcaill \((1,1)\)
\({m_{CP}} = \frac{{ - 2 - 1}}{{5 - 1}} = - \frac{3}{4}\)
Mar sin tha \({m_{tgt}} = \frac{4}{3}\) bhon a tha \({m_{CP}} \times {m_{tgt}} = - 1\)
Mar sin 's e co-aontar an tansaint aig P:
\(y - ( - 2) = \frac{4}{3}(x - 5)\)
\(3(y + 2) = 4(x - 5)\)
\(3y - 4x + 26 = 0\)
\(4x - 3y - 26 = 0\)
Question
Obraich a-mach co-aontar an tansaint ris a' chearcall \({x^2} + {y^2} - 2x - 2y - 23 = 0\) aig a' phuing \((5,4)\)
'S e meadhan a' chearcaill \((1,1)\)
\({m_{radius}} = \frac{{4 - 1}}{{5 - 1}} = \frac{3}{4} \Rightarrow {m_{tgt}} = - \frac{4}{3}\)
\(y - 4 = - \frac{4}{3}(x - 5)\)
\(3(y - 4) = - 4(x - 5)\)
\(3y - 12 = - 4x + 20\)
\(3y + 4x = 32\)
\(4x + 3y - 32 = 0\)
Question
Obraich a-mach co-aontar an tansaint ris a' chearcall \({x^2} + {y^2} - 2x + 5y = 0\) aig a' phuing \((2,0)\)
'S e meadhan a' chearcaill \(\left( {1, - \frac{5}{2}} \right)\)
\({m_{radius}} = \frac{{0 - \left( { - \frac{5}{2}} \right)}}{{2 - 1}} = \frac{5}{2} \Rightarrow {m_{tgt}} = - \frac{2}{5}\)
\(y - 0 = - \frac{2}{5}(x - 2)\)
\(5y = - 2(x - 2)\)
\(5y = - 2x + 4\)
\(5y + 2x = 4\)
\(2x + 5y - 4 = 0\)
