3 szkoły ponadpodstawowej

III liceum

III technikum

Prosto do matury 3. Zakres podstawowy i rozszerzony

Rozwiązanie 1

Wyznaczamy środki boków: MAB​=(25+1​,2−1+5​)=(26​,24​)=(3,2)   MBC​=(21+(−3)​,25+1​)=(21−3​,26​)=(−22​,3)=(−1,3)

Najpierw wyznaczymy współrzędne punktu C. BC<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.522em' viewBox='0 0 400000 522' preserveAspectRatio='xMaxYMin slice'><path d='M0 241v40h399891c-47.3 35.3-84 78-110 128
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 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7
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Odległość między punktami A=(xA​,yA​)  i B=(xB​,yB​)  obliczamy

Szukamy punktu przecięcia prostych  x+y=0  i  2x+y−2=0 {x+y=0  ∣−y2x+y−2=0​   {x=−y2⋅(−y)+y−2=0​   {x=−y−2y+y−2=0​   {x=−y−y−2=0   ∣+y​   {x=−y−2=y​   {x=−yy=−2​   {x=−(−2)y=−2​   {x=2y=−2​ Jednym z wierzchołków trójkąta

Boki tego równoległoboku mają długość ∣AB∣=∣CD∣=(5−0)2+(2−7)2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.88em' viewBox='0 0 400000 1944' preserveAspectRatio='xMinYMin slice'><path d='M983 90
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Środkowe to odcinki łączące wierzchołek ze środkiem naprzeciwległego boku. Najpierw znajdziemy wierzchołek A.

2.18.