3 szkoły ponadpodstawowej

III liceum

III technikum

Prosto do matury 3. Zakres podstawowy i rozszerzony

Skorzystamy ze wzoru na środek ciężkości trójkąta <table style="width:419px;border-collapse:collapse;text-align:center;background-color:#ef6c00;height:68px;" border="0"> <tbody> <tr style="height:106px;"> <td style="width:415px;height:106px;"> S=(3xA​+xB​+xC​​,3yA​+yB​+yC​​)  </td> </tr> </tbody> </table>

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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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

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Środkowe to odcinki łączące wierzchołek ze środkiem naprzeciwległego boku. Najpierw znajdziemy wierzchołek A.

2.15.