find the area of triangle ABC whose vertices are A(-3, -5),B(5,2),C( -9,-3)

We have, A-3,-5; B5,2; C-9,-3 as the vertices of ABC.We know that area of  having its vertices x1,y1; x2,y2; x3,y3 isArea = 12x1y2-y3 + x2y3 - y1 + x3y1-y2Now, area of ABC = 12-32+3 + 5-3+5 - 9-5-2=12-15+10+63=12×58=29 square units

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