1 Elemente de geometrie analitica Cum aflam ecuatia unei drepte ...
1 Elemente de geometrie analitica Cum aflam ecuatia unei drepte ...
1 Elemente de geometrie analitica Cum aflam ecuatia unei drepte ...
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<strong>Cum</strong> <strong>aflam</strong> <strong>ecuatia</strong> <strong>unei</strong> <strong>drepte</strong> ?<br />
<strong>Elemente</strong> <strong>de</strong> <strong>geometrie</strong> <strong>analitica</strong><br />
• daca cunoastem doua puncte <strong>de</strong> pe dreapta A(xA, yA) si B(xB, yB) atunci <strong>ecuatia</strong> <strong>drepte</strong>i<br />
va fi :<br />
d :<br />
x − xA<br />
= y − yA<br />
xB − xA<br />
yB − yA<br />
Exemplu : dreapta care contine punctele A(2, 3) si B(2, 6) are <strong>ecuatia</strong><br />
AB :<br />
x − 2<br />
2 − 2<br />
= y − 3<br />
6 − 3<br />
⇔ x − 2 = 0<br />
daca unul dintre numitori se anuleaza atunci vom anula si numaratorul !<br />
• daca cunoastem un punct <strong>de</strong> pe dreapta A(xA, yA) si panta m atunci <strong>ecuatia</strong> <strong>drepte</strong>i va<br />
fi :<br />
<strong>Cum</strong> <strong>aflam</strong> panta <strong>unei</strong> <strong>drepte</strong> ?<br />
d : y − yA = m(x − xA)<br />
• daca cunoastem doua puncte <strong>de</strong> pe dreapta A(xA, yA) si B(xB, yB) atunci panta <strong>drepte</strong>i<br />
va fi :<br />
mAB = yB − yA<br />
xB − xA<br />
• daca cunoastem <strong>ecuatia</strong> <strong>drepte</strong>i , <strong>de</strong> exemplu :<br />
atunci panta va fi : md = − b<br />
a<br />
d : ax + by + c = 0<br />
• cel mai a<strong>de</strong>sea extragem panta din relatii <strong>de</strong> paralelism si perpendicularitate intre <strong>drepte</strong>:<br />
daca d1 d2 atunci md1 = md2, daca d1 ⊥ d2 atunci md1 ·md2 = −1<br />
<strong>Cum</strong> <strong>aflam</strong> coordonatele unui punct ?<br />
• daca e punctul <strong>de</strong> intersectie a doua <strong>drepte</strong> (sau a <strong>unei</strong> <strong>drepte</strong> cu un cerc, <strong>de</strong> ex) se<br />
rezolva sistemul format cu ecuatiile celor doua <strong>drepte</strong> ( a <strong>drepte</strong>i si a cercului)<br />
• daca stim raportul care-l <strong>de</strong>termina pe un segment cu extremitatile cunoscute A(xA, yA)<br />
si B(xB, yB) avem doua situatii posibile :<br />
daca MA<br />
MB<br />
= k si M este in interiorul segmentului, atunci :<br />
xM = xA+k·xB<br />
1+k<br />
yM = yA+k·yB<br />
1+k<br />
1
2<br />
daca MA<br />
MB<br />
= k si M este in exteriorul segmentului, atunci :<br />
xM = xA−k·xB<br />
1−k<br />
yM = yA−k·yB<br />
1−k<br />
Apartenenta : un punct apartine <strong>unei</strong> <strong>drepte</strong> ( <strong>unei</strong> conice) daca ale sale coordonate<br />
verifica <strong>ecuatia</strong> <strong>drepte</strong>i (conicei)<br />
Formule utile :<br />
• distanta <strong>de</strong> la un punct M(xM, yM) la o dreapta d : ax + by + c = 0 este :<br />
dist(M, d) = |axM + byM + c|<br />
√ a 2 + b 2<br />
• aria triunghiului <strong>de</strong>terminat <strong>de</strong> A(xA, yA), B(xB, yB) si C(xC, yC) :<br />
S∆ABC = |D|<br />
2<br />
<br />
<br />
<br />
un<strong>de</strong> D = <br />
<br />
<br />
xA yA 1<br />
xB yB 1<br />
xC yC 1<br />
• coordonatele mijlocului M a unui segment [AB] : xM = xA+xB<br />
2<br />
• coordonatele centrului <strong>de</strong> greutate G al unui triunghi ∆ABC :<br />
xG = xA+xB+xC<br />
3<br />
si yG = yA+yB+yC<br />
3<br />
• trei puncte A, B, C sunt coliniare daca si numai daca :<br />
<br />
<br />
<br />
<br />
<br />
<br />
xA yA 1<br />
xB yB 1<br />
xC yC 1<br />
<br />
<br />
<br />
<br />
= 0<br />
<br />
• <strong>ecuatia</strong> cercului care trece prin trei puncte A,B si C necoliniare :<br />
cerc :<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
x 2 + y 2 x y 1<br />
x 2 A + y2 A xA yA 1<br />
x 2 B + y2 B xB yB 1<br />
x 2 C + y2 C xC yC 1<br />
<br />
<br />
<br />
<br />
<br />
= 0<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
si yM = yA+yB<br />
2<br />
• <strong>ecuatia</strong> tangentei la cercul x 2 + y 2 + mx + ny + p = 0 in punctul A(xA, yA) se obtine<br />
prin <strong>de</strong>dublare :<br />
x · xA + y · yA + m<br />
2 (x + xA) + n<br />
2 (y + yA) + p = 0<br />
• restul formulelor necesare pentru aplicatiile <strong>de</strong> la cursul <strong>de</strong> recuperare se gasesc in Ghid<br />
<strong>de</strong> recapitulare matematica-C. Lazureanu sau in orice manual <strong>de</strong> clasa a 11-a <strong>de</strong> <strong>geometrie</strong><br />
<strong>analitica</strong>.